Ch02

January 10, 2001 13:09 g65-ch2 Sheet number 1 Page number 107 cyan magenta yellow black
LIMITS AND
CONTINUITY
The problem of defining and calculating instantaneous rates
such as speed and acceleration attracted almost all the
mathematicians of the seventeenth century.
—Morris Kline
he development of calculus in the seventeenth cen-
tury by Newton and Leibniz provided scientists with their
first real understanding of what is meant by an “instanta-
neous rate of change” such as velocity and acceleration.
Once the idea was understood conceptually, efficient com-
putational methods followed, and science took a quantum
leap forward. The fundamental building block on which
rates of change rest is the concept of a “limit,” an idea that
is so important that all other calculus concepts are now
based on it.
In this chapter we will develop the concept of a limit in
stages, proceeding from an informal, intuitive notion to a
precise mathematical definition. We will also develop the-
orems and procedures for calculating limits, and we will
conclude the chapter by using the limits to study “contin-
uous” curves.

108 Limits and Continuity
2.1 LIMITS (AN INTUITIVE APPROACH)
The concept of a limit is the fundamental building block on which all other calculus
concepts are based. In this section we will study limits informally, with the goal of
developing an “intuitive feel” for the basic ideas. In the following three sections we
will focus on the computational methods and precise definitions.
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INSTANTANEOUS VELOCITY AND
THE SLOPE OF A CURVE
Recall from Formula (11) of Section 1.5 that if a particle moves along an s-axis, then the
average velocity vave over the time interval from t0 to t1 is defined as
vave =
s
t
=
s1 − s0
t1 − t0
(1)
where s0 and s1 are the s-coordinates of the particle at times t0 and t1, respectively. Geo-
metrically, vave is the slope of the line joining the points (t0, s0) and (t1, s1) on the position
versus time curve for the particle (Figure 2.1.1).
Slope = v ave
t0 t1
t1 – t0
s1 – s0
t
s
(t0, s0)
(t1, s1)
s = f(t)
Figure 2.1.1
s
0
Figure 2.1.2
Suppose, however, that we are not interested in average velocity over a time interval,
but rather the velocity vinst at a specific instant in time. It is not a simple matter of applying
Formula (1), since the displacement and the elapsed time in an instant are both zero. How-
ever, intuition suggests that the velocity at an instant t = t0 can be approximated by finding
the position of the particle at a time t1 just before, or just after, time t0 and computing the
average velocity over the brief time interval between the two moments. That is,
vinst ≈ vave =
s1 − s0
t1 − t0
(2)
provided t = t1 −t0 is small. Moreover, if we are able to make very precise measurements,
the closer t1 is to t0, the better vave approximates vinst. That is, as we sample at times t1,
closer and closer to t0, vave approaches a limiting value that we understand to be vinst.
Example 1 Suppose that a ball is thrown vertically upward and the height in feet of the
ball t seconds after its release is modeled by the function
s(t) = −16t2
+ 29t + 6, 0 ≤ t ≤ 2
What is a reasonable estimate for the instantaneous velocity of the ball at time t = 0.5 s?
Solution. At any time 0 ≤ t ≤ 2 we may envision the height s(t) of the ball as a position
on a (vertical) s-axis, where s = 0 corresponds to ground level (Figure 2.1.2). The height
of the ball at time t = 0.5 s is s(0.5) = 16.5 ft, and the height of the ball 0.01 s later is
s(0.51) = 16.6284 ft. Therefore, the average velocity of the ball over the time interval from
t = 0.5 to t = 0.51 is
vave =
16.6284 − 16.5
0.51 − 0.5
=
0.1284
0.01
= 12.84 ft/s
Similarly, the height of the ball 0.49 s after its release is s(0.49) = 16.3684 ft, and the
average velocity of the ball over the time interval from t = 0.49 to t = 0.5 is
vave =
16.3684 − 16.5
0.49 − 0.5
=
−0.1316
−0.01
= 13.16 ft/s
Consequently, we would expect the instantaneous velocity of the ball at time t = 0.5 to be
between 12.84 ft/s and 13.16 ft/s. To improve our estimate of this instantaneous velocity,
we can compute the average velocity
vave(t1) =
s(t1) − 16.5
t1 − 0.5
=
−16t2
1 + 29t1 + 6 − 16.5
t1 − 0.5
=
−16t2
1 + 29t1 − 10.5
t1 − 0.5
for values of t1 even closer to 0.5. Table 2.1.1 displays the results of several such computa-

2.1 Limits (An Intuitive Approach) 109
Table 2.1.1
0.5010
0.5005
0.5001
0.4999
0.4995
0.4990
12.9840
12.9920
12.9984
13.0016
13.0080
13.0160
vave(t1) =time t1 (s) (ft/s)
–16t1
2 + 29t1 – 10.5
t1 – 0.5
tions. It appears from these computations that a reasonable estimate for the instantaneous
velocity of the ball at time t = 0.5 s is 13 ft/s.
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FOR THE READER. The domain of the height function s(t) = −16t2
+29t +6 in Example
1 is the closed interval [0, 2]. Why do we not consider values of t less than 0 or greater than
2 for this function? In Table 2.1.1, why is there not a value of vave(t1) for t1 = 0.5?
We can interpret vinst geometrically from the interpretation of vave as the slope of the
line joining the points (t0, s0) and (t1, s1) on the position versus time curve for the particle.
When t = t1 − t0 is small, the points (t0, s0) and (t1, s1) are very close to each other on
the curve. As the sampling point (t1, s1) is selected closer to our anchoring point (t0, s0),
the slope vave more nearly approximates what we might reasonably call the slope of the
position curve at time t = t0. Thus, vinst can be viewed as the slope of the position curve at
time t = t0 (Figure 2.1.3). We will explore this connection more fully in Section 3.1.
Slope = v ave
t0 t1
t
s
Slope=v
inst
Figure 2.1.3
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LIMITS
In Example 1 it appeared that choosing values of t1 close to (but not equal to) 0.5 resulted
in values of vave(t1) that were close to 13. One way of describing this behavior is to say that
the limiting value of vave(t1) as t1 approaches 0.5 is 13 or, equivalently, that 13 is the limit
of vave(t1) as t1 approaches 0.5. More generally, we will see that the concept of the limit of
a function provides a foundation for the tools of calculus. Thus, it is appropriate to start a
study of calculus by focusing on the limit concept itself.
The most basic use of limits is to describe how a function behaves as the independent
variable approaches a given value. For example, let us examine the behavior of the function
f(x) = x2
− x + 1
for x-values closer and closer to 2. It is evident from the graph and table in Figure 2.1.4 that
the values of f(x) get closer and closer to 3 as values of x are selected closer and closer
to 2 on either the left or the right side of 2. We describe this by saying that the “limit of
x2
− x + 1 is 3 as x approaches 2 from either side,” and we write
lim
x →2
(x2
− x + 1) = 3 (3)
Observe that in our investigation of limx →2 (x2
− x + 1) we are only concerned with the
values of f(x) near x = 2 and not the value of f(x) at x = 2.
This leads us to the following general idea.
2.1.1 LIMITS (AN INFORMAL VIEW). If the values of f(x) can be made as close as
we like to L by taking values of x sufﬁciently close to a (but not equal to a), then we
write
lim
x →a
f(x) = L (4)
which is read “the limit of f(x) as x approaches a is L.”

2
3
x
y
xx
f(x)
f(x)
y = f(x) = x2
– x + 1
x
f(x)
1.0
1.000000
1.5
1.750000
1.9
2.710000
1.95
2.852500
1.99
2.970100
1.995
2.985025
1.999
2.997001
2.05
3.152500
2.005
3.015025
2.001
3.003001
2.1
3.310000
2.5
4.750000
3.0
7.000000
2 2.01
3.030100
Left side Right side
Figure 2.1.4
Equation (4) is also commonly written as
f(x)→L as x →a
With this notation we can express (3) as
x2
− x + 1→3 as x →2
In order to investigate limx →a f(x), we ask ourselves the question, “If x is close to,
but different from, a, is there a particular number to which f(x) is close?” This question
presumes that the function f is defined “everywhere near a,” in other words, that f is
defined at all points x in some open interval containing a, except possibly at x = a. The
value of f at a, if it exists at all, is not relevant to the determination of limx →a f(x). Many
important applications of the limit concept involve contexts in which the domain of the
function excludes a. Indeed, our discussion of instantaneous velocity concluded that vinst
could be interpreted as a limit of the average velocities, even though the average velocity
at an instant is not defined.
The process of determining a limit generally involves a discovery phase, followed by
a verification phase. The discovery phase begins with sampled x-values, and ends with
a conjecture for the limit. Figure 2.1.4 illustrates the discovery phase for the problem of
finding the value of limx →2 (x2
− x + 1). We sampled values for x near 2 and found that
the corresponding values of f(x) were close to 3. Indeed, values of x nearer to 2 produced
values of f(x) closer to 3. Our conjecture that limx →2 (x2
− x + 1) = 3 concluded the
discovery phase for this limit. However, a complete treatment of any limit also involves a
verification phase in which it is shown that the conjectured limit is actually correct. For
example, consider our conjecture that limx →2 (x2
− x + 1) = 3. We can only sample a
relatively few values of x near 2, even by using a graphing utility. We cannot sample all
values of x near 2, for no matter how close to 2 we take an x-value, there are infinitely
many values of x nearer yet to 2. To verify that limx →2 (x2
− x + 1) is indeed 3, we need
to resort to an analysis that can overcome this dilemma. This analysis will require a more
mathematically precise definition of limit and is the focus of Section 2.4. In this section,
we concentrate on the discovery phase for limit problems.
Example 2 Make a conjecture about the value of the limit
lim
x →0
x
√
x + 1 − 1
(5)

Solution. Observe that the function
f(x) =
x
√
x + 1 − 1
is not defined at x = 0. However, f is defined for x > −1, x = 0, so the domain of f con-
tains values of x “everywhere near 0.” Table 2.1.2 shows samples of x-values approaching
0 from the left side and from the right side. In both cases the values of f(x), calculated to
six decimal places, appear to get closer and closer to 2, and hence we conjecture that
lim
x →0
x
√
x + 1 − 1
= 2 (6)
A graphing utility could be used to produce Figure 2.1.5, providing more evidence in support
of our conjecture. In the next section we will see that the graph of f(x) is identical to that
of y =
√
x + 1 + 1, except for a hole at (0, 2).
-1 1
x x
1
2
x
y
Figure 2.1.5
Table 2.1.2
–0.01
1.994987
–0.001
1.999500
–0.0001
1.999950
–0.00001
1.999995
0.00001
2.000005
0.0001
2.000050
0.001
2.000500
0.01
2.004988
0x
f(x)
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FOR THE READER. Using a graphing utility, find a window about x = 0 in which all values
of f(x) are within 0.5 of y = 2. Find a window in which all values of f(x) are within 0.1
of y = 2.
Example 3 Make a conjecture about the value of the limit
lim
x →0
sin x
x
(7)
Solution. The function f(x) = (sin x)/x is not defined at x = 0, but, as discussed pre-
viously, this has no bearing on the limit. With the help of a calculating utility set in radian
mode, we obtain the table in Figure 2.1.6.
lim
x →0
sin x
x
= 1 (8)
The result is consistent with the graph of f(x) = (sin x)/x shown in the figure. Later in this
chapter we will give a geometric argument to prove that our conjecture is correct.
1
x 0 x
f(x)
y = f(x) =
sin x
x
As x approaches 0 from the left
or right, f(x) approaches 1.
x
y
±1.0
±0.9
±0.8
±0.7
±0.6
±0.5
±0.4
±0.3
±0.2
±0.1
±0.01
0.84147
0.87036
0.89670
0.92031
0.94107
0.95885
0.97355
0.98507
0.99335
0.99833
0.99998
sin x
xy =
x
(radians)
Figure 2.1.6

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FOR THE READER. Use a calculating utility to sample x-values closer to 0 than in Table ??.
Does the limit change if x is in degrees?
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SAMPLING PITFALLS
Although numerical and graphical evidence is helpful for guessing at limits, we can be
misled by an insufﬁcient or poorly selected sample. For example, the table in Figure 2.1.7
shows values of f(x) = sin(π/x) at selected values of x on both sides of 0. The data
incorrectly suggest that
lim
x →0
sin
π
x
= 0
The fact that this is incorrect is evidenced by the graph of f shown in the ﬁgure. This graph
indicates that as x →0, the values of f oscillate between −1 and 1 with increasing rapidity,
and hence do not approach a limit. The data are deceiving because the table consists only
of sample values of x that are x-intercepts for f(x).
-1 1
-1
1
y = sin ( )x
p
x
y
x = ±1
x = ±0.1
x = ±0.01
x = ±0.001
x = ±0.0001
sin(±p) = 0
sin(±10p) = 0
sin(±100p) = 0
sin(±1000p) = 0
sin(±10,000p) = 0
±p
±10p
±100p
±1000p
±10,000p
x
p
x
p
f(x) = sin ( )
x
(radians)
.
.
.
.
.
.
.
.
.
Figure 2.1.7
Numerical evidence can lead to incorrect conclusions about limits because of roundoff
error or because the sample of values used is not extensive enough to give a good indication
of the behavior of the function. Thus, when a limit is conjectured from a table of values, it
is important to look for corroborating evidence to support the conjecture.
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ONE-SIDED LIMITS
The limit in (4) is commonly called a two-sided limit because it requires the values of f(x)
to get closer and closer to L as values of x are taken from either side of x = a. However,
some functions exhibit different behaviors on the two sides of an x-value a, in which case
it is necessary to distinguish whether values of x near a are on the left side or on the right
side of a for purposes of investigating limiting behavior. For example, consider the function
f(x) =
|x|
x
=
1, x > 0
−1, x < 0
(Figure 2.1.8). Note that x-values approaching 0 and to the right of 0 produce f(x) values
that approach 1 (in fact, they are exactly 1 for all such values of x). On the other hand, x-
values approaching 0 and to the left of 0 produce f(x) values that approach −1. We describe
these two statements by saying that “the limit of f(x) = |x|/x is 1 as x approaches 0 from
the right” and that “the limit of f(x) = |x|/x is −1 as x approaches 0 from the left.” We
denote these limits by writing
lim
x →0+
|x|
x
= 1 and lim
x →0−
|x|
x
= −1 (9–10)
With this notation, the superscript “+” indicates a limit from the right and the superscript
“−” indicates a limit from the left.
This leads to the following general idea.
-1
1
x
y
y =
|x|
x
Figure 2.1.8

2.1.2 ONE-SIDED LIMITS (AN INFORMAL VIEW). If the values of f(x) can be made
as close as we like to L by taking values of x sufficiently close to a (but greater than a),
then we write
lim
x →a+
f(x) = L (11)
which is read “the limit of f(x) as x approaches a from the right is L.” Similarly, if the
values of f(x) can be made as close as we like to L by taking values of x sufficiently
close to a (but less than a), then we write
lim
x →a−
f(x) = L (12)
which is read “the limit of f(x) as x approaches a from the left is L.”
Expressions (11) and (12), which are called one-sided limits, are also commonly written as
f(x)→L as x →a+
and f(x)→L as x →a−
respectively. With this notation (9) and (10) can be expressed as
|x|
x
→1 as x →0+
and
|x|
x
→−1 as x →0−
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THE RELATIONSHIP BETWEEN
ONE-SIDED LIMITS AND
TWO-SIDED LIMITS
In general, there is no guarantee that a function will have a limit at a specified location. If
the values of f(x) do not get closer and closer to some single number L as x → a, then
we say that the limit of f(x) as x approaches a does not exist (and similarly for one-sided
limits). For example, the two-sided limit limx →0 |x|/x does not exist because the values of
f(x) do not approach a single number as x →0; the values approach −1 from the left and
1 from the right.
In general, the following condition must be satisfied for the two-sided limit of a function
to exist.
2.1.3 THE RELATIONSHIP BETWEEN ONE-SIDED AND TWO-SIDED LIMITS. The two-
sided limit of a function f(x) exists at a if and only if both of the one-sided limits exist
at a and have the same value; that is,
lim
x →a
f(x) = L if and only if lim
x →a−
f(x) = L = lim
x →a+
f(x)
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REMARK. Sometimes, one or both of the one-sided limits may fail to exist (which, in
turn, implies that the two-sided limit does not exist). For example, we saw earlier that the
one-sided limits of f(x) = sin(π/x) do not exist as x approaches 0 because the function
keeps oscillating between −1 and 1, failing to settle on a single value. This implies that the
two-sided limit does not exist as x approaches 0.
Example 4 For the functions in Figure 2.1.9, find the one-sided and two-sided limits at
x = a if they exist.
x
y
2
3
1
a
x
y
2
3
1
a
x
y
2
3
1
a
y = f(x) y = f(x) y = f(x)
Figure 2.1.9

Solution. The functions in all three figures have the same one-sided limits as x →a, since
the functions are identical, except at x = a. These limits are
lim
x →a+
f(x) = 3 and lim
x →a−
f(x) = 1
In all three cases the two-sided limit does not exist as x → a because the one-sided limits
are not equal.
Example 5 For the functions in Figure 2.1.10, find the one-sided and two-sided limits
at x = a if they exist.
x
y
2
3
1
a a a
x
y
2
3
1
x
y
2
3
1
y = f(x) y = f(x) y = f(x)
Figure 2.1.10
Solution. As in the preceding example, the value of f at x = a has no bearing on the
limits as x →a, so that in all three cases we have
lim
x →a+
f(x) = 2 and lim
x →a−
f(x) = 2
Since the one-sided limits are equal, the two-sided limit exists and
lim
x →a
f(x) = 2
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INFINITE LIMITS AND VERTICAL
ASYMPTOTES
Sometimes one-sided or two-sided limits will fail to exist because the values of the function
increase or decrease indefinitely. For example, consider the behavior of the function f(x) =
1/x for values of x near 0. It is evident from the table and graph in Figure 2.1.11 that as
x-values are taken closer and closer to 0 from the right, the values of f(x) = 1/x are
positive and increase indefinitely; and as x-values are taken closer and closer to 0 from the
left, the values of f(x) = 1/x are negative and decrease indefinitely. We describe these
x
y
x
y =
1
x
1
x
x→0+
lim = +∞
1
x
x
y
y =
1
x
1
x
x
x→0−
lim = −∞
1
x
–1
–1
–0.1
–10
–0.01
–100
–0.0001
–10,000
0.0001
10,000
0.001
1000
0.01
100
0.1
10
0x –0.001
–1000
1
1
1
x
Figure 2.1.11

limiting behaviors by writing
lim
x →0+
1
x
= +ϱ and lim
x →0−
1
x
= −ϱ
More generally:
2.1.4 INFINITE LIMITS (AN INFORMAL VIEW). If the values of f(x) increase indefi-
nitely as x approaches a from the right or left, then we write
lim
x →a+
f(x) = +ϱ or lim
x →a−
f(x) = +ϱ
as appropriate, and we say that f(x) increases without bound, or f(x) approaches
+ϱ, as x →a+
or as x →a−
. Similarly, if the values of f(x) decrease indefinitely as x
approaches a from the right or left, then we write
lim
x →a+
f(x) = −ϱ or lim
x →a−
f(x) = −ϱ
as appropriate, and say that f(x) decreases without bound, or f(x) approaches −ϱ, as
x →a+
or as x →a−
. Moreover, if both one-sided limits are +ϱ, then we write
lim
x →a
f(x) = +ϱ
and if both one-sided limits are −ϱ, then we write
lim
x →a
f(x) = −ϱ
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REMARK. It should be emphasized that the symbols +ϱ and −ϱ are not real numbers. The
phrase “f(x) approaches +ϱ” is akin to saying that “f(x) approaches the unapproachable”;
it is a colloquialism for “f(x) increases without bound.” The symbols +ϱ and −ϱ are used
here to encapsulate a particular way in which limits fail to exist. To say, for example, that
f(x) → +ϱ as x → a+
is to indicate that limx →a+ f(x) does not exist, and to say further
that this limit fails to exist because values of f(x) increase without bound as x approaches
a from the right. Furthermore, since +ϱ and −ϱ are not numbers, it is inappropriate to
manipulate these symbols using rules of algebra. For example, it is not correct to write
(+ϱ) − (+ϱ) = 0.
Example 6 For the functions in Figure 2.1.12, describe the limits at x = a in appropriate
limit notation.
x
y
x
y
x
y
x
y
1
x – af(x) =
1
(x – a)2
f(x) =
–1
x – af(x) =
–1
(x – a)2
f(x) =
(a) (b) (c) (d)
a a a a
Figure 2.1.12
Solution (a). In Figure 2.1.12a, the function increases indefinitely as x approaches a from
the right and decreases indefinitely as x approaches a from the left. Thus,
lim
x →a+
1
x − a
= +ϱ and lim
x →a−
1
x − a
= −ϱ

Solution (b). In Figure 2.1.12b, the function increases indefinitely as x approaches a from
both the left and right. Thus,
lim
x →a
1
(x − a)2
= lim
x →a+
1
(x − a)2
= lim
x →a−
1
(x − a)2
= +ϱ
Solution (c). In Figure 2.1.12c, the function decreases indefinitely as x approaches a from
the right and increases indefinitely as x approaches a from the left. Thus,
lim
x →a+
−1
x − a
= −ϱ and lim
x →a−
−1
x − a
= +ϱ
Solution (d). In Figure 2.1.12d, the function decreases indefinitely as x approaches a
from both the left and right. Thus,
lim
x →a
−1
(x − a)2
= lim
x →a+
−1
(x − a)2
= lim
x →a−
−1
(x − a)2
= −ϱ
Geometrically, if f(x)→+ϱ as x → a−
or x → a+
, then the graph of y = f(x) rises
without bound and squeezes closer to the vertical line x = a on the indicated side of x = a.
If f(x)→−ϱ as x →a−
or x →a+
, then the graph of y = f(x) falls without bound and
squeezes closer to the vertical line x = a on the indicated side of x = a. In these cases, we
call the line x = a a vertical asymptote. (“Asymptote” comes from the Greek asymptotos,
meaning “nonintersecting.” We will soon see that taking “asymptote” to be synonymous
with “nonintersecting” is a bit misleading.)
2.1.5 DEFINITION. A line x = a is called a vertical asymptote of the graph of a
function f if f(x)→+ϱ or f(x)→−ϱ as x approaches a from the left or right.
Example 7 The four functions graphed in Figure 2.1.12 all have a vertical asymptote at
x = a, which is indicated by the dashed vertical lines in the figure.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS AT INFINITY AND
HORIZONTAL ASYMPTOTES
Thus far, we have used limits to describe the behavior of f(x) as x approaches a. However,
sometimes we will not be concerned with the behavior of f(x) near a specific x-value, but
rather with how the values of f(x) behave as x increases without bound or decreases without
bound. This is sometimes called the end behavior of the function because it describes how
the function behaves for values of x that are far from the origin. For example, it is evident
from the table and graph in Figure 2.1.13 that as x increases without bound, the values of
–10,000
–0.0001
–1000
–0.001
–100
–0.01
–1
–1
–10
–0.1
x decreasing without bound
x
f(x)
1
1
10
0.1
100
0.01
1000
0.001
10,000
0.0001
x increasing without bound
x
f(x)
x
y
x
y =
1
x
1
x
x→+∞
lim = 0
1
x
x
y
y =
1
x
1
x
x
x→−∞
lim = 0
1
x
. . . . . .
. . .. . .
Figure 2.1.13

f(x) = 1/x are positive, but get closer and closer to 0; and as x decreases without bound,
the values of f(x) = 1/x are negative, and also get closer and closer to 0. We indicate these
limiting behaviors by writing
lim
x →+ϱ
1
x
= 0 and lim
x →−ϱ
1
x
= 0
More generally:
2.1.6 LIMITS AT INFINITY (AN INFORMAL VIEW). If the values of f(x) eventually get
closer and closer to a number L as x increases without bound, then we write
lim
x →+ϱ
f(x) = L or f(x)→L as x →+ϱ (13)
Similarly, if the values of f(x) eventually get closer and closer to a number L as x
decreases without bound, then we write
lim
x →−ϱ
f(x) = L or f(x)→L as x →−ϱ (14)
Geometrically, if f(x)→L as x →+ϱ, then the graph of y = f(x) eventually gets
closer and closer to the line y = L as the graph is traversed in the positive direction (Fig-
ure 2.1.14a); and if f(x)→L as x →−ϱ, then the graph of y = f(x) eventually gets
closer and closer to the line y = L as the graph is traversed in the negative x-direction
(Figure 2.1.14b). In either case we call the line y = L a horizontal asymptote of the graph
of f . For example, the function in Figure 2.1.13 all have y = 0 as a horizontal asymptote.
x
y
y = LHorizontal asymptote
x
y
y = LHorizontal asymptote
(a) (b)
Figure 2.1.14
2.1.7 DEFINITION. A line y = L is called a horizontal asymptote of the graph of a
function f if
lim
x →+ϱ
f(x) = L or lim
x →−ϱ
f(x) = L
x
y
y = 3
y =
3x + 1
x
3
Figure 2.1.15
Sometimestheexistenceofahorizontalasymptoteofafunctionf willbereadilyapparent
from the formula for f . For example, it is evident that the function
f(x) =
3x + 1
x
= 3 +
1
x
has a horizontal asymptote at y = 3 (Figure 2.1.15), since the value of 1/x approaches 0 as
x → +ϱ or x → −ϱ. For more complicated functions, algebraic manipulations or special
techniques that we will study in the next section may have to be applied to conﬁrm the
existence of horizontal asymptotes.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
HOW LIMITS AT INFINITY CAN FAIL
TO EXIST
Limits at inﬁnity can fail to exist for various reasons. One possibility is that the values of
f(x) may increase or decrease without bound as x →+ϱ or as x →−ϱ. For example, the
values of f(x) = x3
increase without bound as x → +ϱ and decrease without bound as

x →−ϱ; and for f(x) = −x3
the values decrease without bound as x →+ϱ and increase
without bound as x →−ϱ (Figure 2.1.16). We denote this by writing
lim
x →+ϱ
x3
= +ϱ, lim
x →−ϱ
x3
= −ϱ, lim
x →+ϱ
(−x3
) = −ϱ, lim
x →−ϱ
(−x3
) = +ϱ
x
y
x
y
y = x3
y = –x3
Decreases
without
bound
Decreases
without
bound
Increases
without
bound
Increases
without
bound
Figure 2.1.16
More generally:
2.1.8 INFINITE LIMITS AT INFINITY (AN INFORMAL VIEW). If the values of f(x) in-
crease without bound as x →+ϱ or as x →−ϱ, then we write
lim
x →+ϱ
f(x) = +ϱ or lim
x →−ϱ
f(x) = +ϱ
as appropriate; and if the values of f(x) decrease without bound as x → +ϱ or as
x →−ϱ, then we write
lim
x →+ϱ
f(x) = −ϱ or lim
x →−ϱ
f(x) = −ϱ
as appropriate.
Limits at infinity can also fail to exist because the graph of the function oscillates indef-
initely in such a way that the values of the function do not approach a fixed number and do
not increase or decrease without bound; the trigonometric functions sin x and cos x have
this property, for example (Figure 2.1.17). In such cases we say that the limit fails to exist
because of oscillation.
x
y y = sin x
There is no limit as
x → +∞ or x → –∞.
Figure 2.1.17
EXERCISE SET 2.1 Graphing Calculator C CAS
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1. For the function f graphed in the accompanying figure, find
(a) lim
x →3−
f(x) (b) lim
x →3+
f(x) (c) lim
x →3
f(x)
(d) f(3) (e) lim
x →−ϱ
f(x) (f) lim
x →+ϱ
f(x).
3
x
y
10
y = f(x)
Figure Ex-1
(a) lim
x →2−
f(x) (b) lim
x →2+
f(x) (c) lim
x →2
f(x)
(d) f(2) (e) lim
x →−ϱ
f(x) (f) lim
x →+ϱ
f(x).
2
2
x
y y = f(x)
Figure Ex-2
3. For the function g graphed in the accompanying figure, find
(a) lim
x →4−
g(x) (b) lim
x →4+
g(x) (c) lim
x →4
g(x)
(d) g(4) (e) lim
x →−ϱ
g(x) (f) lim
x →+ϱ
g(x).
4
1
x
y y = g(x)
Figure Ex-3
4. For the function g graphed in the accompanying figure, find
(a) lim
x →0−
g(x) (b) lim
x →0+
g(x) (c) lim
x →0
g(x)
(d) g(0) (e) lim
x →−ϱ
g(x) (f) lim
x →+ϱ
g(x).
4
x
y
5–5
y = g(x)
Figure Ex-4

5. For the function F graphed in the accompanying figure, find
(a) lim
x →−2−
F(x) (b) lim
x →−2+
F(x) (c) lim
x →−2
F(x)
(d) F(−2) (e) lim
x →−ϱ
F(x) (f) lim
x →+ϱ
F(x).
-2
3
x
y y = F(x)
Figure Ex-5
6. For the function F graphed in the accompanying figure, find
(a) lim
x →3−
F(x) (b) lim
x →3+
F(x) (c) lim
x →3
F(x)
(d) F(3) (e) lim
x →−ϱ
F(x) (f) lim
x →+ϱ
F(x).
3
3
x
y y = F(x)
Figure Ex-6
7. For the function φ graphed in the accompanying figure, find
(a) lim
x →−2−
φ(x) (b) lim
x →−2+
φ(x) (c) lim
x →−2
φ(x)
(d) φ(−2) (e) lim
x →−ϱ
φ(x) (f) lim
x →+ϱ
φ(x).
–2
2
x
y y = f(x)
Figure Ex-7
8. For the function φ graphed in the accompanying figure, find
(a) lim
x →4−
φ(x) (b) lim
x →4+
φ(x) (c) lim
x →4
φ(x)
(d) φ(4) (e) lim
x →−ϱ
φ(x) (f) lim
x →+ϱ
φ(x).
4
4
x
y y = f(x)
Figure Ex-8
(a) lim
x →3−
f(x) (b) lim
x →3+
f(x) (c) lim
x →3
f(x)
(d) f(3) (e) lim
x →−ϱ
f(x) (f) lim
x →+ϱ
f(x).
3
4
x
y y = f(x)
Figure Ex-9
(a) lim
x →0−
f(x) (b) lim
x →0+
f(x) (c) lim
x →0
f(x)
(d) f(0) (e) lim
x →−ϱ
f(x) (f) lim
x →+ϱ
f(x).
3
-2
x
y y = f(x)
Figure Ex-10
11. For the function G graphed in the accompanying figure, find
(a) lim
x →0−
G(x) (b) lim
x →0+
G(x) (c) lim
x →0
G(x)
(d) G(0) (e) lim
x →−ϱ
G(x) (f) lim
x →+ϱ
G(x).
1
2
x
y y = G(x)
Figure Ex-11
12. For the function G graphed in the accompanying figure, find
(a) lim
x →0−
G(x) (b) lim
x →0+
G(x) (c) lim
x →0
G(x)
(d) G(0) (e) lim
x →−ϱ
G(x) (f) lim
x →+ϱ
G(x).
4
4
x
y y = G(x)
Figure Ex-12

13. Consider the function g graphed in the accompanying fig-
ure. For what values of x0 does lim
x →x0
g(x) exist?
2–4
2
x
y y = g(x)
Figure Ex-13
14. Consider the function f graphed in the accompanying fig-
ure. For what values of x0 does lim
x →x0
f(x) exist?
3
4
x
y y = f(x)
Figure Ex-14
In Exercises 15–18, sketch a possible graph for a function f
with the specified properties. (Many different solutions are
possible.)
15. (i) f(0) = 2 and f(2) = 1
(ii) lim
x →1−
f(x) = +ϱ and lim
x →1+
f(x) = −ϱ
(iii) lim
x →+ϱ
f(x) = 0 and lim
x →−ϱ
f(x) = +ϱ
16. (i) f(0) = f(2) = 1
(ii) lim
x →2−
f(x) = +ϱ and lim
x →2+
f(x) = 0
(iii) lim
x →−1−
f(x) = −ϱ and lim
x →−1+
f(x) = +ϱ
(iv) lim
x →+ϱ
f(x) = 2 and lim
x →−ϱ
f(x) = +ϱ
17. (i) f(x) = 0 if x is an integer and f(x) = 0 if x is not an
integer
(ii) lim
x →+ϱ
f(x) = 0 and lim
x →−ϱ
f(x) = 0
18. (i) f(x) = 1 if x is a positive integer and f(x) = 1 if
x > 0 is not a positive integer
(ii) f(x) = −1 if x is a negative integer and f(x) = −1
if x < 0 is not a negative integer
(iii) lim
x →+ϱ
f(x) = 1 and lim
x →−ϱ
f(x) = −1
In Exercises 19–22: (i) Make a guess at the limit (if it ex-
ists) by evaluating the function at the specified x-values.
(ii) Confirm your conclusions about the limit by graphing
the function over an appropriate interval. (iii) If you have a
CAS, then use it to find the limit. [Note: For the trigonomet-
ric functions, be sure to set your calculating and graphing
utilities to the radian mode.]
C 19. (a) lim
x →1
x − 1
x3 − 1
; x = 2, 1.5, 1.1, 1.01, 1.001, 0, 0.5, 0.9,
0.99, 0.999
(b) lim
x →1+
x + 1
x3 − 1
; x = 2, 1.5, 1.1, 1.01, 1.001, 1.0001
(c) lim
x →1−
x + 1
x3 − 1
; x = 0, 0.5, 0.9, 0.99, 0.999, 0.9999
C 20. (a) lim
x →0
√
x + 1 − 1
x
; x = ±0.25, ±0.1, ±0.001,
±0.0001
(b) lim
x →0+
√
x + 1 + 1
x
; x = 0.25, 0.1, 0.001, 0.0001
(c) lim
x →0−
√
x + 1 + 1
x
; x = −0.25, −0.1, −0.001,
−0.0001
C 21. (a) lim
x →0
sin 3x
x
; x = ±0.25, ±0.1, ±0.001, ±0.0001
(b) lim
x →−1
cos x
x + 1
; x = 0, −0.5, −0.9, −0.99, −0.999,
−1.5, −1.1, −1.01, −1.001
C 22. (a) lim
x →−1
tan(x + 1)
x + 1
; x = 0, −0.5, −0.9, −0.99, −0.999,
−1.5, −1.1, −1.01, −1.001
(b) lim
x →0
sin(5x)
sin(2x)
; x = ±0.25, ±0.1, ±0.001, ±0.0001
23. Consider the motion of the ball described in Example 1. By
interpreting instantaneous velocity as a limit of average ve-
locity, make a conjecture for the value of the instantaneous
velocity of the ball 0.25 s after its release.
24. Consider the motion of the ball described in Example 1. By
interpreting instantaneous velocity as a limit of average ve-
locity, make a conjecture for the value of the instantaneous
velocity of the ball 0.75 s after its release.
In Exercises 25 and 26: (i) Approximate the y-coordinates
of all horizontal asymptotes of y = f(x) by evaluat-
ing f at the x-values ±10, ±100, ±1000, ±100,000, and
±100,000,000. (ii) Confirm your conclusions by graphing
y = f(x) over an appropriate interval. (iii) If you have a
CAS, then use it to find the horizontal asymptotes.
C 25. (a) f(x) =
2x + 3
x + 4
(b) f(x) = 1 +
3
x
x
(c) f(x) =
x2
+ 1
x + 1

C 26. (a) f(x) =
x2
− 1
5x2 + 1
(b) f(x) = 2 +
1
x
x
(c) f(x) =
sin x
x
27. Assume that a particle is accelerated by a constant force.
The two curves v = n(t) and v = e(t) in the accompanying
figure provide velocity versus time curves for the particle
as predicted by classical physics and by the special theory
of relativity, respectively. The parameter c designates the
speed of light. Using the language of limits, describe the
differences in the long-term predictions of the two theories.
Time
v = n(t)
(Classical)
v = e(t)
(Relativity)
c
Velocity
v
t
Figure Ex-27
28. Let T = f(t) denote the temperature of a baked potato t
minutes after it has been removed from a hot oven. The ac-
companying figure shows the temperature versus time curve
for the potato, where r is the temperature of the room.
(a) What is the physical significance of lim
t →0+
f(t)?
(b) What is the physical significance of lim
t →+ϱ
f(t)?
Time (min)
T = f(t)
Temperature(°F)
T
t
400
r
Figure Ex-28
In Exercises 29 and 30: (i) Conjecture a limit from numerical
evidence. (ii) Use the substitution t = 1/x to express the
limit as an equivalent limit in which t → 0+
or t → 0−
, as
appropriate. (iii) Use a graphing utility to make a conjecture
about your limit in (ii).
29. (a) lim
x →+ϱ
x sin
1
x
(b) lim
x →+ϱ
1 − x
1 + x
(c) lim
x →−ϱ
1 +
2
x
x
30. (a) lim
x →+ϱ
cos(π/x)
π/x
(b) lim
x →+ϱ
x
1 + x
(c) lim
x →−ϱ
(1 − 2x)1/x
31. Suppose that f(x) denotes a function such that
lim
t →0
f(1/t) = L
What can be said about
lim
x →+ϱ
f(x) and lim
x →−ϱ
f(x)?
32. (a) Do any of the trigonometric functions, sin x, cos x,
tan x, cot x, sec x, csc x, have horizontal asymptotes?
(b) Do any of them have vertical asymptotes? Where?
33. (a) Let
f(x) = 1 + x2 1.1/x2
Graph f in the window [−1, 1]×[2.5, 3.5] and use the
calculator’s trace feature to make a conjecture about the
limit of f as x →0.
(b) Graph f in the window [−0.001, 0.001]×[2.5, 3.5] and
use the calculator’s trace feature to make a conjecture
about the limit of f as x →0.
(c) Graph f in the window [−0.000001, 0.000001] ×
[2.5, 3.5] and use the calculator’s trace feature to make
a conjecture about the limit of f as x →0.
(d) Later we will be able to show that
lim
x →0
1 + x2 1.1/x2
≈ 3.00416602
What flaw do your graphs reveal about using numerical
evidence (as revealed by the graphs you obtained) to
make conjectures about limits?
Roundoff error is one source of inaccuracy in calculator
and computer computations. Another source of error, called
catastrophicsubtraction,occurswhentwonearlyequalnum-
bers are subtracted, and the result is used as part of another
calculation. For example, by hand calculation we have
(0.123456789012345 − 0.123456789012344) × 1015
= 1
However, a calculator that can only store 14 decimal digits
produces a value of 0 for this computation, since the num-
bers being subtracted are identical in the first 14 digits. Catas-
trophic subtraction can sometimes be avoided by rearranging
formulas algebraically, but your best defense is to be aware
that it can occur. Watch out for it in the next exercise.
C 34. (a) Let
f(x) =
x − sin x
x3
Make a conjecture about the limit of f as x → 0+
by
evaluating f(x) at x = 0.1, 0.01, 0.001, 0.0001.
(b) Evaluate f(x) at x = 0.000001, 0.0000001,
0.00000001, 0.000000001, 0.0000000001, and make
another conjecture.
(c) What flaw does this reveal about using numerical evi-
dence to make conjectures about limits?
(d) If you have a CAS, use it to show that the exact value
of the limit is 1
6
.

35. (a) The accompanying figure shows two different views of
the graph of the function in Exercise 34, as generated
by Mathematica. What is happening?
(b) Use your graphing utility to generate the graphs, and
see whether the same problem occurs.
(c) Would you expect a similar problem to occur in the
vicinity of x = 0 for the function
f(x) =
1 − cos x
x
?
See if it does.
-0.001 -0.0005 0.0005 0.001
0.166667
0.166667
0.166667
0.166667
0.166667
-0.01 -0.005 0.005 0.01
0.166666
0.166666
0.166666
0.166667
Erratic graph generated by Mathematica
Figure Ex-35
2.2 COMPUTING LIMITS
In this section we will discuss algebraic techniques for computing limits of many func-
tions. We base these results on the informal development of the limit concept discussed
in the preceding section. A more formal derivation of these results is possible after
Section 2.4.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
SOME BASIC LIMITS
Our strategy for finding limits algebraically has two parts:
• First we will obtain the limits of some simple functions.
• Then we will develop a repertoire of theorems that will enable us to use the limits
of those simple functions as building blocks for finding limits of more complicated
functions.
We start with the cases of a constant function f(x) = k, the identity function f(x) = x,
and the reciprocal function f(x) = 1/x.
2.2.1 THEOREM. Let a and k be real numbers.
lim
x →a
k = k lim
x →a
x = a
lim
x →0−
1
x
= −ϱ lim
x →0+
1
x
= +ϱ
The four limits in Theorem 2.2.1 should be evident from inspection of the function graphs
shown in Figure 2.2.1.
In the case of the constant function f(x) = k, the values of f(x) do not change as x
varies, so the limit of f(x) is k, regardless of at which number a the limit is taken. For
example,
lim
x →−25
3 = 3, lim
x →0
3 = 3, lim
x →π
3 = 3

2.2 Computing Limits 123
y = x
x a x
a
f(x) = x
f(x) = x
x
y
x
y
x
y
x
y
x a x
x →a
lim k = k
x →a
lim x = a
y = f(x) = k
k
x
y =
1
xy =
1
x
1
x
1
x
x
x→0+
lim = +∞
1
xx→0−
lim = −∞
1
x
Figure 2.2.1
Since the identity function f(x) = x just echoes its input, it is clear that f(x) = x →a
as x →a. In terms of our informal deﬁnition of limits (2.1.1), if we decide just how close
to a we would like the value of f(x) = x to be, we need only restrict its input x to be just
as close to a.
The one-sided limits of the reciprocal function f(x) = 1/x about 0 should conform
with your experience with fractions: making the denominator closer to zero increases the
magnitude of the fraction (i.e., increases its absolute value). This is illustrated in Table 2.2.1.
Table 2.2.1
values conclusion
–1
–1
1
1
x
1/x
x
1/x
–0.1
–10
0.1
10
– 0.01
–100
0.01
100
–0.001
–1000
0.001
1000
–0.0001
–10,000
0.0001
10,000
. . .
. . .
. . .
. . .
As x → 0–
the value of 1/x
decreases without bound.
As x → 0+
the value of 1/x
increases without bound.
The following theorem, parts of which are proved in Appendix G, will be our basic tool
for ﬁnding limits algebraically.
2.2.2 THEOREM. Let a be a real number, and suppose that
lim
x →a
f(x) = L1 and lim
x →a
g(x) = L2
That is, the limits exist and have values L1 and L2, respectively. Then,
(a) lim
x →a
[f(x) + g(x)] = lim
x →a
f(x) + lim
x →a
g(x) = L1 + L2
(b) lim
x →a
[f(x) − g(x)] = lim
x →a
f(x) − lim
x →a
g(x) = L1 − L2
(c) lim
x →a
[f(x)g(x)] = lim
x →a
f(x) lim
x →a
g(x) = L1L2
(d) lim
x →a
f(x)
g(x)
=
lim
x →a
f(x)
lim
x →a
g(x)
=
L1
L2
, provided L2 = 0
(e) lim
x →a
n
f(x) = n lim
x →a
f(x) = n
L1, provided L1 > 0 if n is even.
Moreover, these statements are also true for one-sided limits.

A casual restatement of this theorem is as follows:
(a) The limit of a sum is the sum of the limits.
(b) The limit of a difference is the difference of the limits.
(c) The limit of a product is the product of the limits.
(d) The limit of a quotient is the quotient of the limits, provided the limit of the denom-
inator is not zero.
(e) The limit of an nth root is the nth root of the limit.
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REMARK. Although results (a) and (c) in Theorem 2.2.2 are stated for two functions, they
hold for any ﬁnite number of functions. For example, if the limits of f(x), g(x), and h(x)
exist as x →a, then the limit of their sum and the limit of their product also exist as x →a
and are given by the formulas
lim
x →a
[f(x) + g(x) + h(x)] = lim
x →a
f(x) + lim
x →a
g(x) + lim
x →a
h(x)
lim
x →a
[f(x)g(x)h(x)] = lim
x →a
f(x) lim
x →a
g(x) lim
x →a
h(x)
In particular, if f(x) = g(x) = h(x), then this yields
lim
x →a
[f(x)]3
= lim
x →a
f(x)
3
More generally, if n is a positive integer, then the limit of the nth power of a function is the
nth power of the function’s limit. Thus,
lim
x →a
xn
= lim
x →a
x
n
= an
(1)
For example,
lim
x →3
x4
= 34
= 81
Another useful result follows from part (c) of Theorem 2.2.2 in the special case when
one of the factors is a constant k:
lim
x →a
(k · f(x)) = lim
x →a
k · lim
x →a
f(x) = k · lim
x →a
f(x) (2)
and similarly for limx →a replaced by a one-sided limit, limx →a+ or limx →a− . Rephrased,
this last statement says:
A constant factor can be moved through a limit symbol.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS OF POLYNOMIALS AND
RATIONAL FUNCTIONS AS x → a
Example 1 Find lim
x →5
(x2
− 4x + 3) and justify each step.
Solution. First note that limx →5 x2
= 52
= 25 by Equation (1). Also, from Equation (2),
limx →5 4x = 4(limx →5 x) = 4(5) = 20. Since limx →5 3 = 3 by Theorem 2.2.1, we may
appeal to Theorem 2.2.2(a) and (b) to write
lim
x →5
(x2
− 4x + 3) = lim
x →5
x2
− lim
x →5
4x + lim
x →5
3 = 25 − 20 + 3 = 8
However, for conciseness, it is common to reverse the order of this argument and simply

write
lim
x →5
(x2
− 4x + 3) = lim
x →5
x2
− lim
x →5
4x + lim
x →5
3 Theorem 2.2.2(a), (b)
= lim
x →5
x
2
− 4 lim
x →5
x + lim
x →5
3 Equations (1), (2)
= 52
− 4(5) + 3 Theorem 2.2.1
= 8
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REMARK. Inourpresentationoflimitarguments,wewilladopttheconventionofproviding
just a concise, reverse argument, bearing in mind that the validity of each equality may be
conditional upon the successful resolution of the remaining limits.
Our next result will show that the limit of a polynomial p(x) at x = a is the same as
the value of the polynomial at x = a. This greatly simpliﬁes the computation of limits of
polynomials by allowing us to simply evaluate the polynomial.
2.2.3 THEOREM. For any polynomial
p(x) = c0 + c1x + · · · + cnxn
and any real number a,
lim
x →a
p(x) = c0 + c1a + · · · + cnan
= p(a)
Proof.
lim
x →a
p(x) = lim
x →a
c0 + c1x + · · · + cnxn
= lim
x →a
c0 + lim
x →a
c1x + · · · + lim
x →a
cnxn
= lim
x →a
c0 + c1 lim
x →a
x + · · · + cn lim
x →a
xn
= c0 + c1a + · · · + cnan
= p(a)
Recall that a rational function is a ratio of two polynomials. Theorem 2.2.3 and Theorem
2.2.2(d) can often be used in combination to compute limits of rational functions.
Example 2 Find lim
x →2
5x3
+ 4
x − 3
.
Solution.
lim
x →2
5x3
+ 4
x − 3
=
lim
x →2
(5x3
+ 4)
lim
x →2
(x − 3)
Theorem 2.2.2(d )
=
5 · 23
+ 4
2 − 3
= −44 Theorem 2.2.3
2.2.4 THEOREM. Consider the rational function
f(x) =
n(x)
d(x)
where n(x) and d(x) are polynomials. For any real number a,
(a) if d(a) = 0, then lim
x →a
f(x) = f(a).
(b) if d(a) = 0 but n(a) = 0, then lim
x →a
f(x) does not exist.

Proof. If d(a) = 0, then
lim
x →a
f(x) = lim
x →a
n(x)
d(x)
=
lim
x →a
n(x)
lim
x →a
d(x)
Theorem 2.2.2(d )
=
n(a)
d(a)
= f(a) Theorem 2.2.3
If d(a) = 0 and n(a) = 0, then we again appeal to your experience with fractions. For
values of x sufﬁciently near a, the value of n(x) will be near n(a) and not zero. Thus, since
0 = d(a) = limx →a d(x), as values of x approach a, the magnitude (absolute value) of the
fraction n(x)/d(x) will increase without bound, so limx →a f(x) does not exist.
As an illustration of part (b) of Theorem 2.2.4, consider
lim
x →3
5x3
+ 4
x − 3
Note that limx →3(5x3
+ 4) = 5 · 33
+ 4 = 139 and limx →3(x − 3) = 3 − 3 = 0. It is
evident from Table 2.2.2 that
lim
x →3
5x3
+ 4
x − 3
does not exist.
Table 2.2.2
values conclusion
2.99
–13,765.45
2.999
–138,865.04
2.9999
–1,389,865.00
. . .
. . .
3.01
14,035.45
3.001
139,135.05
3.0001
1,390,135.00
x
5x3 + 4
x – 3
5x3 + 4
x – 3
5x3 + 4
x – 3
x
5x3 + 4
x – 3
. . .
. . .
The value of decreases
without bound as x → 3–
.
The value of increases
without bound as x → 3+
.
In Theorem 2.2.4(b), where the limit of the denominator is zero but the limit of the
numerator is not zero, the response “does not exist” can be elaborated upon in one of the
following three ways.
• The limit may be −ϱ.
• The limit may be +ϱ.
• The limit may be −ϱ from one side and +ϱ from the other.
Figure 2.2.2 illustrates these three possibilities graphically for rational functions of the form
1/(x − a), 1/(x − a)2
, and −1/(x − a)2
.
Example 3 Find
(a) lim
x →4−
2 − x
(x − 4)(x + 2)
(b) lim
x →4+
2 − x
(x − 4)(x + 2)
(c) lim
x →4
2 − x
(x − 4)(x + 2)
Solution. With n(x) = 2 − x and d(x) = (x − 4)(x + 2), we see that n(4) = −2 and
d(4) = 0. By Theorem 2.2.4(b), each of the limits does not exist. To be more speciﬁc, we

x xx
a a a
y =
1
x – a
y = 1
(x – a)2
y = – 1
(x – a)2
1
x – ax→a+
lim = +∞
1
x – ax→a–
lim = −∞
1
(x – a)2
x→a
lim = +∞
1
(x – a)2
x→a
lim − = −∞
Figure 2.2.2
analyze the sign of the ratio n(x)/d(x) near x = 4. The sign of the ratio, which is given
in Figure 2.2.3, is determined by the signs of 2 − x, x − 4, and x + 2. (The method of
test values, discussed in Appendix A, provides a simple way of finding the sign of the ratio
here.) It follows from this figure that as x approaches 4 from the left, the ratio is always
positive; and as x approaches 4 from the right, the ratio is always negative. Thus,
lim
x →4−
2 − x
(x − 4)(x + 2)
= +ϱ and lim
x →4+
2 − x
(x − 4)(x + 2)
= −ϱ
Because the one-sided limits have opposite signs, all we can say about the two-sided limit
is that it does not exist.
–2 2 4
0+ + + – – – – – – –+ +
Sign of
2 − x
(x − 4)(x + 2)
x
Figure 2.2.3
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
INDETERMINATE FORMS OF TYPE
0/0
The missing case in Theorem 2.2.4 is when both the numerator and the denominator of a
rational function f(x) = n(x)/d(x) have a zero at x = a. In this case, n(x) and d(x) will
each have a factor of x − a, and canceling this factor may result in a rational function to
which Theorem 2.2.4 applies.
Example 4 Find lim
x →2
x2
− 4
x − 2
.
Solution. Since 2 is a zero of both the numerator and denominator, they share a common
factor of x − 2. The limit can be obtained as follows:
lim
x →2
x2
− 4
x − 2
= lim
x →2
(x − 2)(x + 2)
x − 2
= lim
x →2
(x + 2) = 4
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REMARK. Although correct, the second equality in the preceding computation needs some
justification, since canceling the factor x − 2 alters the function by expanding its domain.
However, as discussed in Example 5 of Section 1.2, the two functions are identical, except at
x = 2 (Figure 1.2.9). From our discussions in the last section, we know that this difference
has no effect on the limit as x approaches 2.
Example 5 Find
(a) lim
x →3
x2
− 6x + 9
x − 3
(b) lim
x →−4
2x + 8
x2 + x − 12
(c) lim
x →5
x2
− 3x − 10
x2 − 10x + 25
Solution (a). The numerator and the denominator both have a zero at x = 3, so there is a
common factor of x − 3. Then,
lim
x →3
x2
− 6x + 9
x − 3
= lim
x →3
(x − 3)2
x − 3
= lim
x →3
(x − 3) = 0

Solution (b). The numerator and the denominator both have a zero at x = −4, so there is
a common factor of x − (−4) = x + 4. Then,
lim
x →−4
2x + 8
x2 + x − 12
= lim
x →−4
2(x + 4)
(x + 4)(x − 3)
= lim
x →−4
2
x − 3
= −
2
7
Solution (c). The numerator and the denominator both have a zero at x = 5, so there is a
common factor of x − 5. Then,
lim
x →5
x2
− 3x − 10
x2 − 10x + 25
= lim
x →5
(x − 5)(x + 2)
(x − 5)(x − 5)
= lim
x →5
x + 2
x − 5
However,
lim
x →5
(x + 2) = 7 = 0 and lim
x →5
(x − 5) = 0
By Theorem 2.2.4(b),
lim
x →5
x2
− 3x − 10
x2 − 10x + 25
= lim
x →5
x + 2
x − 5
does not exist.
The case of a limit of a quotient,
lim
x →a
f(x)
g(x)
where limx →a f(x) = 0 and limx →a g(x) = 0, is called an indeterminate form of type
0/0. Note that the limits in Examples 4 and 5 produced a variety of answers. The word
“indeterminate” here refers to the fact that the limiting behavior of the quotient cannot
be determined without further study. The expression “0/0” is just a mnemonic device
to describe the circumstance of a limit of a quotient in which both the numerator and
denominator approach 0.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS INVOLVING RADICALS
Example 6 Find lim
x →0
x
√
x + 1 − 1
.
Solution. Recall that in Example 2 of Section 2.1 we conjectured this limit to be 2. Note
that this limit expression is an indeterminate form of type 0/0, so Theorem 2.2.2(d) does
not apply. One strategy for resolving this limit is to first rationalize the denominator of the
function. This yields
x
√
x + 1 − 1
=
x(
√
x + 1 + 1)
(x + 1) − 1
=
√
x + 1 + 1, x = 0
Therefore,
lim
x →0
x
√
x + 1 − 1
= lim
x →0
(
√
x + 1 + 1) = 2
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS OF PIECEWISE-DEFINED
FUNCTIONS
For functions that are defined piecewise, a two-sided limit at an x-value where the formula
changes is best obtained by first finding the one-sided limits at that number.
Example 7 Let
f(x) =



1/(x + 2), x < −2
x2
− 5, −2 < x ≤ 3
√
x + 13, x > 3
Find
(a) lim
x →−2
f(x) (b) lim
x →0
f(x) (c) lim
x →3
f(x)

Solution (a). As x approaches −2 from the left, the formula for f is
f(x) =
1
x + 2
so that
lim
x →2−
f(x) = lim
x →2−
1
x + 2
= −ϱ
As x approaches −2 from the right, the formula for f is
f(x) = x2
− 5
so that
lim
x →−2+
f(x) = lim
x →2+
(x2
− 5) = (−2)2
− 5 = −1
Thus, limx →−2 f(x) does not exist.
Solution (b). As x approaches 0 from either the left or the right, the formula for f is
f(x) = x2
− 5
Thus,
lim
x →0
f(x) = lim
x →0
(x2
− 5) = 02
− 5 = −5
Solution (c). As x approaches 3 from the left, the formula for f is
f(x) = x2
− 5
so that
lim
x →3−
f(x) = lim
x →3−
(x2
− 5) = 32
− 5 = 4
As x approaches 3 from the right, the formula for f is
f(x) =
√
x + 13
so that
lim
x →3+
f(x) = lim
x →3+
√
x + 13 = lim
x →3+
(x + 13) =
√
3 + 13 = 4
Since the one-sided limits are equal, we have
lim
x →3
f(x) = 4
EXERCISE SET 2.2
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1. In each part, find the limit by inspection.
(a) lim
x →8
7 (b) lim
x →0+
π
(c) lim
x →−2
3x (d) lim
y →3+
12y
2. In each part, find the stated limit of f(x) = x/|x| by in-
spection.
(a) lim
x →5
f(x) (b) lim
x →−5
f(x)
(c) lim
x →0+
f(x) (d) lim
x →0−
f(x)
3. Given that
lim
x →a
f(x) = 2, lim
x →a
g(x) = −4, lim
x →a
h(x) = 0
find the limits that exist. If the limit does not exist, explain
why.
(a) lim
x →a
[f(x) + 2g(x)] (b) lim
x →a
[h(x) − 3g(x) + 1]
(c) lim
x →a
[f(x)g(x)] (d) lim
x →a
[g(x)]2
(e) lim
x →a
3
6 + f(x) (f) lim
x →a
2
g(x)
(g) lim
x →a
3f(x) − 8g(x)
h(x)
(h) lim
x →a
7g(x)
2f(x) + g(x)
4. Use the graphs of f and g in the accompanying figure to
why.

(a) lim
x →2
[f(x) + g(x)] (b) lim
x →0
[f(x) + g(x)]
(c) lim
x →0+
[f(x) + g(x)] (d) lim
x →0−
[f(x) + g(x)]
(e) lim
x →2
f(x)
1 + g(x)
(f) lim
x →2
1 + g(x)
f(x)
(g) lim
x →0+
f(x) (h) lim
x →0−
f(x)
1
1
x
y
1
1
x
y
y = f(x) y = g(x)
Figure Ex-4
In Exercises 5–30, find the limits.
5. lim
y →2−
(y − 1)(y − 2)
y + 1
6. lim
x →3
x2
− 2x
x + 1
7. lim
x →4
x2
− 16
x − 4
8. lim
x →0
6x − 9
x3 − 12x + 3
9. lim
x →1+
x4
− 1
x − 1
10. lim
t →−2
t3
+ 8
t + 2
11. lim
x →−1
x2
+ 6x + 5
x2 − 3x − 4
12. lim
x →2
x2
− 4x + 4
x2 + x − 6
13. lim
t →2
t3
+ 3t2
− 12t + 4
t3 − 4t
14. lim
t →1
t3
+ t2
− 5t + 3
t3 − 3t + 2
15. lim
x →3+
x
x − 3
16. lim
x →3−
x
x − 3
17. lim
x →3
x
x − 3
18. lim
x →2+
x
x2 − 4
19. lim
x →2−
x
x2 − 4
20. lim
x →2
x
x2 − 4
21. lim
y →6+
y + 6
y2 − 36
22. lim
y →6−
y + 6
y2 − 36
23. lim
y →6
y + 6
y2 − 36
24. lim
x →4+
3 − x
x2 − 2x − 8
25. lim
x →4−
3 − x
x2 − 2x − 8
26. lim
x →4
3 − x
x2 − 2x − 8
27. lim
x →2+
1
|2 − x|
28. lim
x →3−
1
|x − 3|
29. lim
x →9
x − 9
√
x − 3
30. lim
y →4
4 − y
2 −
√
y
31. Verify the limit in Example 1 of Section 2.1. That is, find
lim
t1 →0.5
−16t2
1 + 29t1 − 10.5
t1 − 0.5
32. Let s(t) = −16t2
+ 29t + 6. Find
lim
t →1.5
s(t) − s(1.5)
t − 1.5
33. Let
f(x) =
x − 1, x ≤ 3
3x − 7, x > 3
Find
(a) lim
x →3−
f(x) (b) lim
x →3+
f(x) (c) lim
x →3
f(x).
34. Let
g(t) =
t2
, t ≥ 0
t − 2, t < 0
Find
(a) lim
t →0−
g(t) (b) lim
t →0+
g(t) (c) lim
t →0
g(t).
35. Let f(x) =
x3
− 1
x − 1
.
(a) Find lim
x →1
f(x).
(b) Sketch the graph of y = f(x).
36. Let
f(x) =



x2
− 9
x + 3
, x = −3
k, x = −3
(a) Find k so that f (−3) = lim
x →−3
f (x).
(b) With k assigned the value limx →−3 f (x), show that
f (x) can be expressed as a polynomial.
37. (a) Explain why the following calculation is incorrect.
lim
x →0+
1
x
−
1
x2
= lim
x →0+
1
x
− lim
x →0+
1
x2
= +ϱ − (+ϱ) = 0
(b) Show that lim
x →0+
1
x
−
1
x2
= −ϱ.
38. Find lim
x →0−
1
x
+
1
x2
.
In Exercises 39 and 40, first rationalize the numerator, then
find the limit.
39. lim
x →0
√
x + 4 − 2
x
40. lim
x →0
x2 + 4 − 2
x
41. Let p(x) and q(x) be polynomials, and suppose q(x0) = 0.
Discuss the behavior of the graph of y = p(x)/q(x) in the
vicinity of x = x0. Give examples to support your conclu-
sions.

2.3 Computing Limits: End Behavior 131
2.3 COMPUTING LIMITS: END BEHAVIOR
In this section we will discuss algebraic techniques for computing limits at ±ϱ for
many functions. We base these results on the informal development of the limit concept
discussed in Section 2.1. A more formal development of these results is possible after
Section 2.4.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
SOME BASIC LIMITS
The behavior of a function toward the extremes of its domain is sometimes called its end
behavior. Here we will use limits to investigate the end behavior of a function as x →−ϱ or
as x →+ϱ. As in the last section, we will begin by obtaining limits of some simple functions
and then use these as building blocks for ﬁnding limits of more complicated functions.
2.3.1 THEOREM. Let k be a real number.
lim
x →−ϱ
k = k lim
x →+ϱ
k = k
lim
x →−ϱ
x = −ϱ lim
x →+ϱ
x = +ϱ
lim
x →−ϱ
1
x
= 0 lim
x →+ϱ
1
x
= 0
The six limits in Theorem 2.3.1 should be evident from inspection of the function graphs
in Figure 2.3.1.
x →−∞
lim x = −∞ x →+∞
lim x = +∞
y = x
x
f(x) = x
y = x
x
f(x) = x
x
y
x
y
x
y
x
y
x
y
x x
k
y = f(x) = k
x → +∞
lim k = k, lim k = k
x → −∞
y =
1
x
1
x
y =
1
x
1
x
x
x
x→+∞
lim = 0
1
xx→−∞
lim = 0
1
x
Figure 2.3.1

The limits of the reciprocal function f (x) = 1/x should make sense to you intuitively,
based on your experience with fractions: increasing the magnitude of x makes its reciprocal
closer to zero. This is illustrated in Table 2.3.1.
Table 2.3.1
values conclusion
–1
–1
1
1
x
1/x
x
1/x
–10
–0.1
10
0.1
–100
–0.01
100
0.01
–1000
–0.001
1000
0.001
–10,000
–0.0001
10,000
0.0001
. . .
. . .
. . .
. . .
As x → –∞ the value of 1/x
increases toward zero.
As x → +∞ the value of 1/x
decreases toward zero.
The following theorem mirrors Theorem 2.2.2 as our tool for ﬁnding limits at ±ϱ alge-
braically. (The proof is similar to that of the portions of Theorem 2.2.2 that are proved in
Appendix G.)
2.3.2 THEOREM. Suppose that
lim
x →+ϱ
f(x) = L1 and lim
x →+ϱ
g(x) = L2
That is, the limits exist and have values L1 and L2, respectively. Then,
(a) lim
x →+ϱ
[f(x) + g(x)] = lim
x →+ϱ
f(x) + lim
x →+ϱ
g(x) = L1 + L2
(b) lim
x →+ϱ
[f(x) − g(x)] = lim
x →+ϱ
f(x) − lim
x →+ϱ
g(x) = L1 − L2
(c) lim
x →+ϱ
[f(x)g(x)] = lim
x →+ϱ
f(x) lim
x →+ϱ
g(x) = L1L2
(d) lim
x →+ϱ
f(x)
g(x)
=
lim
x →+ϱ
f(x)
lim
x →+ϱ
g(x)
=
L1
L2
, provided L2 = 0
(e) lim
x →+ϱ
n
f(x) = n lim
x →+ϱ
f(x) = n
L1, provided L1 > 0 if n is even.
Moreover, these statements are also true if x →−ϱ.
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REMARK. As in the remark following Theorem 2.2.2, results (a) and (c) can be extended to
sums or products of any ﬁnite number of functions. In particular, for any positive integer n,
lim
x →+ϱ
(f(x))n
= lim
x →+ϱ
f(x)
n
lim
x →−ϱ
(f(x))n
= lim
x →−ϱ
f(x)
n
Also, since limx →+ϱ(1/x) = 0, if n is a positive integer, then
lim
x →+ϱ
1
xn
= lim
x →+ϱ
1
x
n
= 0 lim
x →−ϱ
1
xn
= lim
x →−ϱ
1
x
n
= 0 (1)
For example,
lim
x →+ϱ
1
x4
= 0 and lim
x →−ϱ
1
x4
= 0
Another useful result follows from part (c) of Theorem 2.3.2 in the special case where
one of the factors is a constant k:
lim
x →+ϱ
(k · f(x)) = lim
x →+ϱ
k · lim
x →+ϱ
f(x) = k · lim
x →+ϱ
f(x) (2)

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and similarly, for limx →+ϱ replaced by limx →−ϱ. Rephrased, this last statement says:
A constant factor can be moved through a limit symbol.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS OF xn AS x → ±∞
In Figure 2.3.2 we have graphed the polynomials of the form xn
for n = 1, 2, 3, and 4.
Below each ﬁgure we have indicated the limits as x →+ϱ and as x →−ϱ. The results in
the ﬁgure are special cases of the following general results:
lim
x →+ϱ
xn
= +ϱ, n = 1, 2, 3, . . . (3)
lim
x →−ϱ
xn
=
−ϱ, n = 1, 3, 5, . . .
+ϱ, n = 2, 4, 6, . . .
(4)
-4 4
-8
8
y = x
x→+∞
lim x = +∞
x→−∞
lim x = −∞
x→+∞
lim x2
= +∞
x→−∞
lim x2
= +∞
x→+∞
lim x4
= +∞
x→−∞
lim x4
= +∞
x→+∞
lim x3
= +∞
x→−∞
lim x3
= −∞
-4 4
-8
8
y = x2
-4 4
-8
8 y = x3
-4 4
-8
8 y = x4
x
y
x
y
x
y
x
y
Figure 2.3.2
Multiplying xn
by a positive real number does not affect limits (3) and (4), but multiplying
by a negative real number reverses the sign.
Example 1
lim
x →+ϱ
2x5
= +ϱ, lim
x →−ϱ
2x5
= −ϱ
lim
x →+ϱ
−7x6
= −ϱ, lim
x →−ϱ
−7x6
= −ϱ
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS OF POLYNOMIALS AS
x → ±∞
There is a useful principle about polynomials which, expressed informally, states that:
The end behavior of a polynomial matches the end behavior of its highest degree term.
More precisely, if cn = 0 then
lim
x →−ϱ
c0 + c1x + · · · + cnxn
= lim
x →−ϱ
cnxn
(5)
lim
x →+ϱ
c0 + c1x + · · · + cnxn
= lim
x →+ϱ
cnxn
(6)
We can motivate these results by factoring out the highest power of x from the polynomial

and examining the limit of the factored expression. Thus,
c0 + c1x + · · · + cnxn
= xn c0
xn
+
c1
xn−1
+ · · · + cn
As x →−ϱ or x →+ϱ, it follows from (1) that all of the terms with positive powers of x
in the denominator approach 0, so (5) and (6) are certainly plausible.
Example 2
lim
x →−ϱ
(7x5
− 4x3
+ 2x − 9) = lim
x →−ϱ
7x5
= −ϱ
lim
x →−ϱ
(−4x8
+ 17x3
− 5x + 1) = lim
x →−ϱ
−4x8
= −ϱ
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
LIMITS OF RATIONAL FUNCTIONS
AS x → ±∞
Ausefultechniquefordeterminingtheendbehaviorofarationalfunctionf(x) = n(x)/d(x)
is to factor and cancel the highest power of x that occurs in the denominator d(x) from
both n(x) and d(x). The denominator of the resulting fraction then has a (nonzero) limit
equal to the leading coefficient of d(x), so the limit of the resulting fraction can be quickly
determined using (1), (5), and (6). The following examples illustrate this technique.
Example 3 Find lim
x →+ϱ
3x + 5
6x − 8
.
Solution. Divide the numerator and denominator by the highest power of x that occurs
in the denominator; that is, x1
= x. We obtain
lim
x →+ϱ
3x + 5
6x − 8
= lim
x →+ϱ
x(3 + 5/x)
x(6 − 8/x)
= lim
x →+ϱ
3 + 5/x
6 − 8/x
=
lim
x →+ϱ
(3 + 5/x)
lim
x →+ϱ
(6 − 8/x)
=
lim
x →+ϱ
3 + lim
x →+ϱ
5/x
lim
x →+ϱ
6 − lim
x →+ϱ
8/x
=
3 + 5 lim
x →+ϱ
1/x
6 − 8 lim
x →+ϱ
1/x
=
3 + (5 · 0)
6 − (8 · 0)
=
1
2
Example 4 Find
(a) lim
x →−ϱ
4x2
− x
2x3 − 5
(b) lim
x →−ϱ
5x3
− 2x2
+ 1
3x + 5
Solution (a). Divide the numerator and denominator by the highest power of x that occurs
in the denominator, namely x3
. We obtain
lim
x →−ϱ
4x2
− x
2x3 − 5
= lim
x →−ϱ
x3
(4/x − 1/x2
)
x3(2 − 5/x3)
= lim
x →−ϱ
4/x − 1/x2
2 − 5/x3
=
lim
x →−ϱ
(4/x − 1/x2
)
lim
x →−ϱ
(2 − 5/x3
)
=
(4 · 0) − 0
2 − (5 · 0)
=
0
2
= 0
Solution (b). Divide the numerator and denominator by x to obtain
lim
x →−ϱ
5x3
− 2x2
+ 1
3x + 5
= lim
x →−ϱ
5x2
− 2x + 1/x
3 + 5/x
= +ϱ
where the final step is justified by the fact that
5x2
− 2x →+ϱ,
1
x
→0, and 3 +
5
x
→3
as x →−ϱ.

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LIMITS INVOLVING RADICALS
Example 5 Find lim
x →+ϱ
3 3x + 5
6x − 8
.
Solution.
lim
x →+ϱ
3 3x + 5
6x − 8
=
3
lim
x →+ϱ
3x + 5
6x − 8
Theorem 2.3.2(e)
=
3 1
2
Example 3
Example 6 Find
(a) lim
x →+ϱ
x2 + 2
3x − 6
(b) lim
x →−ϱ
x2 + 2
3x − 6
In both parts it would be helpful to manipulate the function so that the powers of x are
transformed to powers of 1/x. This can be achieved in both cases by dividing the numerator
and denominator by |x| and using the fact that
√
x2 = |x|.
Solution (a). As x → +ϱ, the values of x under consideration are positive, so we can
replace |x| by x where helpful. We obtain
lim
x →+ϱ
x2 + 2
3x − 6
= lim
x →+ϱ
x2 + 2/|x|
(3x − 6)/|x|
= lim
x →+ϱ
x2 + 2/
√
x2
(3x − 6)/x
= lim
x →+ϱ
1 + 2/x2
3 − 6/x
=
lim
x →+ϱ
1 + 2/x2
lim
x →+ϱ
(3 − 6/x)
=
lim
x →+ϱ
(1 + 2/x2)
lim
x →+ϱ
(3 − 6/x)
=
lim
x →+ϱ
1 + 2 lim
x →+ϱ
1/x2
lim
x →+ϱ
3 − 6 lim
x →+ϱ
1/x
=
1 + (2 · 0)
3 − (6 · 0)
=
1
3
Solution (b). As x → −ϱ, the values of x under consideration are negative, so we can
replace |x| by −x where helpful. We obtain
lim
x →−ϱ
x2 + 2
3x − 6
= lim
x →−ϱ
x2 + 2/|x|
(3x − 6)/|x|
= lim
x →−ϱ
x2 + 2/
√
x2
(3x − 6)/(−x)
= lim
x →−ϱ
1 + 2/x2
−3 + 6/x
= −
1
3
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FOR THE READER. Use a graphing utility to explore the end behavior of
f(x) =
√
x2 + 2
3x − 6
Your investigation should support the results of Example 6.
-2 -1 1 2 3 4
-1
1
2
3
4
x
y
y = √x6
+ 5 – x3
-1 1 2 3 4
-1
1
2
3
4
x
y
y = 2.5
y = √x6
+ 5x3
– x3
, x ≥ 0
(a)
(b)
Figure 2.3.3
Example 7 Find
(a) lim
x →+ϱ
( x6 + 5 − x3
) (b) lim
x →+ϱ
( x6 + 5x3 − x3
)
Solution. Graphs of the functions f(x) =
√
x6 + 5−x3
and g(x) =
√
x6 + 5x3 −x3
for
x ≥ 0 are shown in Figure 2.3.3. From the graphs we might conjecture that the limits are 0
and 2.5, respectively. To conﬁrm this, we treat each function as a fraction with denominator

1 and rationalize the numerator.
lim
x →+ϱ
( x6 + 5 − x3
) = lim
x →+ϱ
( x6 + 5 − x3
)
√
x6 + 5 + x3
√
x6 + 5 + x3
= lim
x →+ϱ
(x6
+ 5) − x6
√
x6 + 5 + x3
= lim
x →+ϱ
5
√
x6 + 5 + x3
= lim
x →+ϱ
5/x3
√
1 + 5/x6 + 1
√
x6 = x3
for x > 0
=
0
√
1 + 0 + 1
= 0
lim
x →+ϱ
( x6 + 5x3 − x3
) = lim
x →+ϱ
( x6 + 5x3 − x3
)
√
x6 + 5x3 + x3
√
x6 + 5x3 + x3
= lim
x →+ϱ
(x6
+ 5x3
) − x6
√
x6 + 5x3 + x3
= lim
x →+ϱ
5x3
√
x6 + 5x3 + x3
= lim
x →+ϱ
5
√
1 + 5/x3 + 1
√
x6 = x3
for x > 0
=
5
√
1 + 0 + 1
=
5
2
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REMARK. Example 7 illustrates an indeterminate form of type ∞ – ∞. Exercises 31–34
explore more examples of this type.
EXERCISE SET 2.3 Graphing Calculator
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1. In each part, ﬁnd the limit by inspection.
(a) lim
x →−ϱ
(−3) (b) lim
h→+ϱ
(−2h)
2. In each part, ﬁnd the stated limit of f(x) = x/|x| by in-
spection.
(a) lim
x →+ϱ
f(x) (b) lim
x →−ϱ
f(x)
3. Given that
lim
x →+ϱ
f(x) = 3, lim
x →+ϱ
g(x) = −5, lim
x →+ϱ
h(x) = 0
why.
(a) lim
x →+ϱ
[f(x) + 3g(x)] (b) lim
x →+ϱ
[h(x) − 4g(x) + 1]
(c) lim
x →+ϱ
[f(x)g(x)] (d) lim
x →+ϱ
[g(x)]2
(e) lim
x →+ϱ
3
5 + f(x) (f) lim
x →+ϱ
3
g(x)
(g) lim
x →+ϱ
3h(x) + 4
x2
(h) lim
x →+ϱ
6f(x)
5f(x) + 3g(x)
4. Given that
lim
x →−ϱ
f(x) = 7, lim
x →−ϱ
g(x) = −6
why.
(a) lim
x →−ϱ
[2f(x) − g(x)] (b) lim
x →−ϱ
[6f(x) + 7g(x)]
(c) lim
x →−ϱ
[x2
+ g(x)] (d) lim
x →−ϱ
[x2
g(x)]
(e) lim
x →−ϱ
3
f(x)g(x) (f) lim
x →−ϱ
g(x)
f(x)
(g) lim
x →−ϱ
f(x) +
g(x)
x
(h) lim
x →−ϱ
xf(x)
(2x + 3)g(x)
5. lim
x →−ϱ
(3 − x) 6. lim
x →−ϱ
5 −
1
x
7. lim
x →+ϱ
(1 + 2x − 3x5
) 8. lim
x →+ϱ
(2x3
−100x+5)
9. lim
x →+ϱ
√
x 10. lim
x →−ϱ
√
5 − x
11. lim
x →+ϱ
3x + 1
2x − 5
12. lim
x →+ϱ
5x2
− 4x
2x2 + 3
13. lim
y →−ϱ
3
y + 4
14. lim
x →+ϱ
1
x − 12
15. lim
x →−ϱ
x − 2
x2 + 2x + 1
16. lim
x →+ϱ
5x2
+ 7
3x2 − x
17. lim
x →+ϱ
3 2 + 3x − 5x2
1 + 8x2
18. lim
s →+ϱ
3 3s7 − 4s5
2s7 + 1

19. lim
x →−ϱ
5x2 − 2
x + 3
20. lim
x →+ϱ
5x2 − 2
x + 3
21. lim
y →−ϱ
2 − y
7 + 6y2
22. lim
y →+ϱ
2 − y
7 + 6y2
23. lim
x →−ϱ
3x4 + x
x2 − 8
24. lim
x →+ϱ
3x4 + x
x2 − 8
25. lim
x →+ϱ
7 − 6x5
x + 3
26. lim
t →−ϱ
5 − 2t3
t2 + 1
27. lim
t →+ϱ
6 − t3
7t3 + 3
28. lim
x →−ϱ
x + 4x3
1 − x2 + 7x3
29. Let
f(x) =



2x2
+ 5, x < 0
3 − 5x3
1 + 4x + x3
, x ≥ 0
Find
(a) lim
x →−ϱ
f(x) (b) lim
x →+ϱ
f(x).
30. Let
g(t) =



2 + 3t
5t2 + 6
, t < 1,000,000
√
36t2 − 100
5 − t
, t > 1,000,000
Find
(a) lim
t →−ϱ
g(t) (b) lim
t →+ϱ
g(t).
31. lim
x →+ϱ
( x2 + 3 − x) 32. lim
x →+ϱ
( x2 − 3x − x)
33. lim
x →+ϱ
( x2 + ax − x)
34. lim
x →+ϱ
( x2 + ax − x2 + bx)
35. Discuss the limits of p(x) = (1 − x)n
as x → +ϱ and
x →−ϱ for positive integer values of n.
36. Let p(x) = (1 − x)n
and q(x) = (1 − x)m
. Discuss the
limits of p(x)/q(x) as x → +ϱ and x → −ϱ for positive
integer values of m and n.
37. Let p(x) be a polynomial of degree n. Discuss the limits
of p(x)/xm
as x → +ϱ and x → −ϱ for positive integer
values of m.
38. In each part, find examples of polynomials p(x) and q(x)
that satisfy the stated condition and such that p(x) → +ϱ
and q(x)→+ϱ as x →+ϱ.
(a) lim
x →+ϱ
p(x)
q(x)
= 1 (b) lim
x →+ϱ
p(x)
q(x)
= 0
(c) lim
x →+ϱ
p(x)
q(x)
= +ϱ (d) lim
x →+ϱ
[p(x) − q(x)] = 3
39. Assuming that m and n are positive integers, find
lim
x →−ϱ
2 + 3xn
1 − xm
[Hint: Your answer will depend on whether m < n, m = n,
or m > n.]
40. Find
lim
x →+ϱ
c0 + c1x + · · · + cnxn
d0 + d1x + · · · + dmxm
where cn = 0 and dm = 0. [Hint: Your answer will depend
on whether m < n, m = n, or m > n.]
The notion of an asymptote can be extended to include curves
as well as lines. Specifically, we say that f(x) is asymptotic
to g(x) as x → +∞ if
lim
x →+ϱ
[f(x) − g(x)] = 0
and that f(x) is asymptotic to g(x) as x → –∞ if
lim
x →−ϱ
[f(x) − g(x)] = 0
Informally stated, if f(x) is asymptotic to g(x) as x →+ϱ,
then the graph of y = f(x) gets closer and closer to the graph
of y = g(x) as x →+ϱ, and if f(x) is asymptotic to g(x) as
x → −ϱ, then the graph of y = f(x) gets closer and closer
to the graph of y = g(x) as x →−ϱ. For example, if
f(x) = x2
+
2
x − 1
and g(x) = x2
then f(x) is asymptotic to g(x) as x →+ϱ and as x →−ϱ
since
lim
x →+ϱ
[f(x) − g(x)] = lim
x →+ϱ
1
x − 1
= 0
lim
x →−ϱ
[f(x) − g(x)] = lim
x →−ϱ
1
x − 1
= 0
This asymptotic behavior is illustrated in the following figure,
which also shows the vertical asymptote of f(x) at x = 1.
-4 -3 -2 -1 2 3 4
-10
-5
5
10
15
20
x
y
y = f(x)
y = g(x)
In Exercises 41–46, determine a function g(x) to which f(x)
is asymptotic as x →+ϱ or x →−ϱ. Use a graphing utility
to generate the graphs of y = f(x) and y = g(x) and identify
all vertical asymptotes.
41. f(x) =
x2
− 2
x − 2
42. f(x) =
x3
− x + 3
x
43. f(x) =
−x3
+ 3x2
+ x − 1
x − 3
44. f(x) =
x5
− x3
+ 3
x2 − 1
45. f(x) = sin x +
1
x − 1
46. f(x) =
x3 − x2 + 2
x − 1

2.4 LIMITS (DISCUSSED MORE RIGOROUSLY)
Thus far, our discussion of limits has been based on our intuitive feeling of what it
means for the values of a function to get closer and closer to a limiting value. How-
ever, this level of informality can only take us so far, so our goal in this section is to
define limits precisely. From a purely mathematical point of view these definitions are
needed to establish limits with certainty and to prove theorems about them. However,
they will also provide us with a deeper understanding of the limit concept, making it
possible for us to visualize some of the more subtle properties of functions.
InSections2.1to2.3ouremphasiswasonthediscoveryofvaluesoflimits,eitherthroughthe
sampling of selected x-values or through the application of limit theorems. In the preceding
sections we interpreted limx →a f(x) = L to mean that the values of f(x) can be made
as close as we like to L by selecting x-values sufficiently close to a (but not equal to a).
Although this informal definition is sufficient for many purposes, we need a more precise
definition to verify that a conjectured limit is actually correct, or to prove the limit theorems
in Sections 2.2 and 2.3. One of our goals in this section is to give the informal phrases “as
close as we like to L” and “sufficiently close to a” a precise mathematical interpretation.
This will enable us to replace the informal definition of limit given in Definition 2.1.1 with
a more fully developed version that may be used in proofs.
To start, consider the function f graphed in Figure 2.4.1a for which f(x)→L as x →a.
We have intentionally placed a hole in the graph at x = a to emphasize that the function
f need not be defined at x = a to have a limit there. Also, to simplify the discussion, we
have chosen a function that is increasing on an open interval containing a.
a x1 x1x0 x0
L −
L +
L
a x
L −
L +
L
y = f(x)y = f(x)y = f(x)
x
y
x
y
a
L
x
y
(a) (b) (c)
Figure 2.4.1
To motivate an appropriate definition for a two-sided limit, suppose that we choose any
positive number, say , and draw horizontal lines from L + and L − on the y-axis to the
curve y = f(x) and then draw vertical lines from those points on the curve to the x-axis. As
shown in Figure 2.4.1b, let x0 and x1 be points where the vertical lines intersect the x-axis.
Next, imagine that x gets closer and closer to a (from either side). Eventually, x will
lie inside the interval (x0, x1), which is marked in green in Figure 2.4.1c; and when this
happens, the value of f(x) will fall between L − and L + , marked in red in the figure.
Thus, we conclude:
If f(x) → L as x → a, then for any positive number , we can find an open interval
(x0, x1) on the x-axis that contains a and has the property that for each x in that
interval (except possibly for x = a), the value of f(x) is between L − and L + .
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FOR THE READER. Consider the limit, limx →0(sin x)/x, conjectured to be 1 in Example
3 of Section 2.1. Draw a figure similar to Figure 2.4.1 that illustrates the preceding analysis
for this limit.

2.4 Limits (Discussed More Rigorously) 139
What is important about this result is that it holds no matter how small we make .
However, making smaller and smaller forces f(x) closer and closer to L—which is
precisely the concept we were trying to capture mathematically.
Observe that in Figure 2.4.1c the interval (x0, x1) extends farther on the right side of a
than on the left side. However, for many purposes it is preferable to have an interval that
extends the same distance on both sides of a. For this purpose, let us choose any positive
number δ that is smaller than both x1 −a and a −x0, and consider the interval (a −δ, a +δ).
This interval extends the same distance δ on both sides of a and lies inside of the interval
(x0, x1) (Figure 2.4.2). Moreover, the condition L − < f(x) < L + holds for every
x in this interval (except possibly x = a), since this condition holds on the larger interval
(x0, x1). This is illustrated by graphing f in the window (a −δ, a +δ)×(L− , L+ ) and
observing that the graph “exits” the window at the sides, not at the top or bottom (except
possibly at x = a).
a – d a + d
a – d a + d
ax0 x1
x
( ( ((
d d
x1x0 a
L −
L +
L
y = f(x)
x
y
Figure 2.4.2
Example 1 Let f(x) = 1
2
x + 1
4
sin(πx/2). It can be shown that lim
x →1
f(x) = L = 0.75.
Let = 0.05.
(a) Use a graphing utility to find an open interval (x0, x1) containing a = 1 such that for
each x in this interval, f(x) is between L − = 0.75 − = 0.75 − 0.05 = 0.70 and
L + = 0.75 + = 0.75 + 0.05 = 0.80.
(b) Find a value of δ such that f(x) is between 0.70 and 0.80 for every x in the interval
(1 − δ, 1 + δ).
Solution (a). Figure 2.4.3 displays the graph of f . With a graphing utility, we discover
that (to five decimal places) the points (0.90769, 0.70122) and (1.09231, 0.79353) are
on the graph of f . Suppose that we take x0 = 0.908 and x1 = 1.09. Since the graph
of f rises from left to right, we see that for x0 = 0.908 < x < 1.090 = x1, we have
0.90769 < x < 1.09231 and therefore 0.7 < 0.70122 < f(x) < 0.79353 < 0.8.
Solution (b). Since x1 −a = 1.09−1 = 0.09 and a −x0 = 1−0.908 = 0.902, any value
or δ that is less than 0.09 will be acceptable. For example, for δ = 0.08, if x belongs to the
interval (1 − δ, 1 + δ) = (0.92, 1.08), then f(x) will lie between 0.70 and 0.80.
0 0.5 1
0
0.5
1
x
y
y = x + sin( )1
2
px
2
1
4
Figure 2.4.3
Note that the condition L − < f(x) < L + can be expressed as
|f(x) − L| <
and the condition that x lies in the interval (a − δ, a + δ), but x = a, can be expressed as
0 < |x − a| < δ
Thus, we can summarize this discussion in the following definition.
2.4.1 LIMIT DEFINITION. Let f(x) be defined for all x in some open interval con-
taining the number a, with the possible exception that f(x) need not be defined at a. We
will write
lim
x →a
f(x) = L
if given any number > 0 we can find a number δ > 0 such that
|f(x) − L| < if 0 < |x − a| < δ
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REMARK. With this definition we have made the transition from informal to formal in
the definition of a two-sided limit. The phrase “as close as we like to L” has been given
quantitative meaning by the number > 0, and the phrase “sufficiently close to a” has been

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made precise by the number δ > 0. Commonly known as the “ -δ definition” of a limit,
Definition 2.4.1 was developed primarily by the German mathematician Karl Weierstrass
∗
in the nineteenth century.
The definitions for one-sided limits are similar to Definition 2.4.1. For example, in the
definition of limx →a+ f(x) we assume that f(x) is defined for all x in an interval of the
form (a, b) and replace the condition 0 < |x − a| < δ by the condition a < x < a + δ.
Comparable changes are made in the definition of limx →a− f(x).
In the preceding sections we illustrated various numerical and graphical methods for
guessing at limits. Now that we have a precise definition to work with, we can actually
confirm the validity of those guesses with mathematical proof. Here is a typical example of
such a proof.
Example 2 Use Definition 2.4.1 to prove that lim
x →2
(3x − 5) = 1.
Solution. We must show that given any positive number , we can find a positive number
δ such that
| (3x − 5)
f(x)
− 1
L
| < if 0 < |x − 2
a
| < δ (1)
There are two things to do. First, we must discover a value of δ for which this statement
holds, and then we must prove that the statement holds for that δ. For the discovery part we
begin by simplifying (1) and writing it as
|3x − 6| < if 0 < |x − 2| < δ
Next, we will rewrite this statement in a form that will facilitate the discovery of an appro-
priate δ:
3|x − 2| < if 0 < |x − 2| < δ
|x − 2| < /3 if 0 < |x − 2| < δ
(2)
It should be self-evident that this last statement holds if δ = /3, which completes the
discovery portion of our work. Now we need to prove that (1) holds for this choice of δ.
However, statement (1) is equivalent to (2), and (2) holds with δ = /3, so (1) also holds
with δ = /3. This proves that limx →2 (3x − 5) = 1.
∗
KARL WEIERSTRASS (1815–1897). Weierstrass, the son of a customs officer, was born in Ostenfelde, Germany.
As a youth Weierstrass showed outstanding skills in languages and mathematics. However, at the urging of his
dominant father, Weierstrass entered the law and commerce program at the University of Bonn. To the chagrin of
his family, the rugged and congenial young man concentrated instead on fencing and beer drinking. Four years
later he returned home without a degree. In 1839 Weierstrass entered the Academy of Münster to study for a career
in secondary education, and he met and studied under an excellent mathematician named Christof Gudermann.
Gudermann’s ideas greatly influenced the work of Weierstrass. After receiving his teaching certificate, Weierstrass
spent the next 15 years in secondary education teaching German, geography, and mathematics. In addition, he
taught handwriting to small children. During this period much of Weierstrass’s mathematical work was ignored
because he was a secondary schoolteacher and not a college professor. Then, in 1854, he published a paper of
major importance that created a sensation in the mathematics world and catapulted him to international fame
overnight. He was immediately given an honorary Doctorate at the University of Königsberg and began a new
career in college teaching at the University of Berlin in 1856. In 1859 the strain of his mathematical research
caused a temporary nervous breakdown and led to spells of dizziness that plagued him for the rest of his life.
Weierstrass was a brilliant teacher and his classes overflowed with multitudes of auditors. In spite of his fame,
he never lost his early beer-drinking congeniality and was always in the company of students, both ordinary and
brilliant. Weierstrass was acknowledged as the leading mathematical analyst in the world. He and his students
opened the door to the modern school of mathematical analysis.

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REMARK. This example illustrates the general form of a limit proof: We assume that we
are given a positive number , and we try to prove that we can find a positive number δ such
that
|f(x) − L| < if 0 < |x − a| < δ (3)
This is done by first discovering δ, and then proving that the discovered δ works. Since
the argument has to be general enough to work for all positive values of , the quantity
δ has to be expressed as a function of . In Example 2 we found the function δ = /3
by some simple algebra; however, most limit proofs require a little more algebraic and
logical ingenuity. Thus, if you find our ensuing discussion of “ -δ” proofs challenging, do
not become discouraged; the concepts and techniques are intrinsically difficult. In fact, a
precise understanding of limits evaded the finest mathematical minds for more than 150
years after the basic concepts of calculus were discovered.
Example 3 Prove that lim
x →0+
√
x = 0.
Solution. Note that the domain of
√
x is 0 ≤ x, so it is valid to discuss the limit as x →0+
.
We must show that given > 0, there exists a δ > 0 such that
|
√
x − 0| < if 0 < x < 0 + δ
or more simply,
√
x < if 0 < x < δ (4)
But, by squaring both sides of the inequality
√
x < , we can rewrite (4) as
x < 2
if 0 < x < δ (5)
It should be self-evident that (5) is true if δ = 2
; and since (5) is a reformulation of (4), we
have shown that (4) holds with δ = 2
. This proves that limx →0+
√
x = 0.
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REMARK. In this example the limit from the left and the two-sided limit do not exist at
x = 0 because the domain of
√
x includes no numbers to the left of 0.
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THE VALUE OF δ IS NOT UNIQUE
In preparation for our next example, we note that the value of δ in Definition 2.4.1 is not
unique; once we have found a value of δ that fulfills the requirements of the definition, then
any smaller positive number δ1 will also fulfill those requirements. That is, if it is true that
|f(x) − L| < if 0 < |x − a| < δ
then it will also be true that
|f(x) − L| < if 0 < |x − a| < δ1
This is because {x : 0 < |x − a| < δ1} is a subset of {x : 0 < |x − a| < δ} (Figure 2.4.4),
and hence if |f(x) − L| < is satisfied for all x in the larger set, then it will automatically
be satisfied for all x in the subset. Thus, in Example 2, where we used δ = /3, we could
have used any smaller value of δ such as δ = /4, δ = /5, or δ = /6.
a – d1 a + d1a
x
(
a – d
( (
a + d
(
Figure 2.4.4
x →3
x2
= 9.
Solution. We must show that given any positive number , we can find a positive number
δ such that
|x2
− 9| < if 0 < |x − 3| < δ (6)
Because |x −3| occurs on the right side of this “if statement,” it will be helpful to factor the
left side to introduce a factor of |x − 3|. This yields the following alternative form of (6)
|x + 3||x − 3| < if 0 < |x − 3| < δ (7)

Using the triangle inequality, we see that
|x + 3| = |(x − 3) + 6| ≤ |x − 3| + 6
Therefore, if 0 < |x − 3| < δ then
|x + 3||x − 3| ≤ (|x − 3| + 6)|x − 3| < (δ + 6)δ
It follows that (7) will be satisfied for any positive value of δ such that (δ + 6)δ ≤ . Let
us agree to restrict our attention to positive values of δ such that δ ≤ 1. (This is justified
because of our earlier observation that once a value of δ is found, then any smaller positive
value of δ can be used.) With this restriction, (δ + 6)δ ≤ 7δ, so that (7) will be satisfied as
long as it is also the case that 7δ ≤ . We can achieve this by taking δ to be the minimum
of the numbers /7 and 1, which is sometimes written as δ = min( /7, 1). This proves that
limx →3 x2
= 9.
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REMARK. You may have wondered how we knew to make the restriction δ ≤ 1 (as opposed
to δ ≤ 1
2
or δ ≤ 5, for example). Actually, it does not matter; any restriction of the form
δ ≤ c would work equally well.
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LIMITS AS x → ±∞
In Section 2.1 we discussed the limits
lim
x →+ϱ
f(x) = L and lim
x →−ϱ
f(x) = L
from an intuitive viewpoint. We interpreted the first statement to mean that the values of
f(x) eventually get closer and closer to L as x increases indefinitely, and we interpreted the
second statement to mean that the values of f(x) eventually get closer and closer to L as x
decreases indefinitely. These ideas are captured more precisely in the following definitions
and are illustrated in Figure 2.4.5.
2.4.2 DEFINITION. Let f(x) be defined for all x in some infinite open interval ex-
tending in the positive x-direction. We will write
lim
x →+ϱ
f(x) = L
if given any number > 0, there corresponds a positive number N such that
|f(x) − L| < if x > N
2.4.3 DEFINITION. Let f(x) be defined for all x in some infinite open interval ex-
tending in the negative x-direction. We will write
lim
x →−ϱ
f(x) = L
if given any number > 0, there corresponds a negative number N such that
|f(x) − L| < if x < N
To see how these definitions relate to our informal concepts of these limits, suppose
that f(x) → L as x → +ϱ, and for a given let N be the positive number described in
Definition 2.4.2. If x is allowed to increase indefinitely, then eventually x will lie in the
interval (N, +ϱ), which is marked in green in Figure 2.4.5a; when this happens, the value
of f(x) will fall between L − and L + , marked in red in the figure. Since this is true for
all positive values of (no matter how small), we can force the values of f(x) as close as
we like to L by making N sufficiently large. This agrees with our informal concept of this
limit. Similarly, Figure 2.4.5b illustrates Definition 2.4.3.

N
L − e
L + e
L
| f(x) − L| < e if x > N | f(x) − L| < e if x < N
N
L − e
L + e
L
xx
yy
f(x) f(x)
(a) (b)
x x
Figure 2.4.5
x →+ϱ
1
x
= 0.
Solution. Applying Definition 2.4.2 with f(x) = 1/x and L = 0, we must show that
given > 0, we can find a number N > 0 such that
1
x
− 0 < if x > N (8)
Because x →+ϱ we can assume that x > 0. Thus, we can eliminate the absolute values in
this statement and rewrite it as
1
x
< if x > N
or, on taking reciprocals,
x >
1
if x > N (9)
It is self-evident that N = 1/ satisfies this requirement, and since (9) is equivalent to (8)
for x > 0, the proof is complete.
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INFINITE LIMITS
In Section 2.1 we discussed limits of the following type from an intuitive viewpoint:
lim
x →a
f(x) = +ϱ, lim
x →a
f(x) = −ϱ (10)
lim
x →a+
f(x) = +ϱ, lim
x →a+
f(x) = −ϱ (11)
lim
x →a−
f(x) = +ϱ, lim
x →a−
f(x) = −ϱ (12)
Recall that each of these expressions describes a particular way in which the limit fails to
exist. The +ϱ indicates that the limit fails to exist because f(x) increases without bound,
and the −ϱ indicates that the limit fails to exist because f(x) decreases without bound.
These ideas are captured more precisely in the following definitions and are illustrated in
Figure 2.4.6.
2.4.4 DEFINITION. Let f(x) be defined for all x in some open interval containing a,
except that f(x) need not be defined at a. We will write
lim
x →a
f(x) = +ϱ
if given any positive number M, we can find a number δ > 0 such that f(x) satisfies
f(x) > M if 0 < |x − a| < δ

x
y
a – d a + d
M
x
y
a – d a + d
M
f(x) > M if 0 < |x − a| < d f(x) < M if 0 < |x − a| < d
(a) (b)
Figure 2.4.6
2.4.5 DEFINITION. Let f(x) be defined for all x in some open interval containing a,
except that f(x) need not be defined at a. We will write
lim
x →a
f(x) = −ϱ
if given any negative number M, we can find a number δ > 0 such that f(x) satisfies
f(x) < M if 0 < |x − a| < δ
To see how these definitions relate to our informal concepts of these limits, suppose
that f(x) → +ϱ as x → a, and for a given M let δ be the corresponding positive number
described in Definition 2.4.4. Next, imagine that x gets closer and closer to a (from either
side). Eventually, x will lie in the interval (a − δ, a + δ), which is marked in green in
Figure 2.4.6a; when this happens the value of f(x) will be greater than M, marked in red in
the figure. Since this is true for any positive value of M (no matter how large), we can force
the values of f(x) to be as large as we like by making x sufficiently close to a. This agrees
with our informal concept of this limit. Similarly, Figure 2.4.6b illustrates Definition 2.4.5.
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REMARK. Thedefinitionsfortheone-sidedlimitsaresimilar.Forexample,inthedefinition
of limx →a− f(x) = +ϱ we assume that f(x) is defined for all x in some interval of the
form (c, a) and replace the condition 0 < |x − a| < δ by the condition a − δ < x < a.
x →0
1
x2
= +ϱ.
Solution. Applying Definition 2.4.4 with f(x) = 1/x2
and a = 0, we must show that
given a number M > 0, we can find a number δ > 0 such that
1
x2
> M if 0 < |x − 0| < δ (13)
or, on taking reciprocals and simplifying,
x2
<
1
M
if 0 < |x| < δ (14)
But x2
< 1/M if |x| < 1/
√
M, so that δ = 1/
√
M satisfies (14). Since (13) is equivalent
to (14), the proof is complete.
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FOR THE READER. How would you define
lim
x →+ϱ
f(x) = +ϱ, lim
x →+ϱ
f(x) = −ϱ
lim
x →−ϱ
f(x) = +ϱ, lim
x →−ϱ
f(x) = −ϱ?
(15)

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1. (a) Find the largest open interval, centered at the origin on
the x-axis, such that for each x in the interval the value
of the function f(x) = x + 2 is within 0.1 unit of the
number f(0) = 2.
(b) Find the largest open interval, centered at x = 3, such
that for each x in the interval the value of the func-
tion f(x) = 4x − 5 is within 0.01 unit of the number
f(3) = 7.
(c) Find the largest open interval, centered at x = 4, such
that for each x in the interval the value of the func-
tion f(x) = x2
is within 0.001 unit of the number
f(4) = 16.
2. In each part, find the largest open interval, centered at
x = 0, such that for each x in the interval the value of
f(x) = 2x + 3 is within units of the number f(0) = 3.
(a) = 0.1 (b) = 0.01
(c) = 0.0012
3. (a) Find the values of x1 and x2 in the accompanying figure.
(b) Find a positive number δ such that |
√
x − 2| < 0.05 if
0 < |x − 4| < δ.
4x1 x2
2 – 0.05
2 + 0.05
2
x
y
Not drawn to scale
y = √x
Figure Ex-3
(b) Find a positive number δ such that |(1/x) − 1| < 0.1 if
0 < |x − 1| < δ.
1
1 – 0.1
1 + 0.1
1
x
y
x1
x2
Not drawn to scale
y =
1
x
Figure Ex-4
5. Generate the graph of f(x) = x3
− 4x + 5 with a graph-
ing utility, and use the graph to find a number δ such that
|f(x) − 2| < 0.05 if 0 < |x − 1| < δ. [Hint: Show
that the inequality |f(x) − 2| < 0.05 can be rewritten as
1.95 < x3
− 4x + 5 < 2.05, and estimate the values of x
for which x3
− 4x + 5 = 1.95 and x3
− 4x + 5 = 2.05.]
6. Use the method of Exercise 5 to find a number δ such that
|
√
5x + 1 − 4| < 0.5 if 0 < |x − 3| < δ.
7. Let f(x) = x +
√
x with L = limx →1 f(x) and let = 0.2.
Use a graphing utility and its trace feature to find a positive
number δ such that |f(x) − L| < if 0 < |x − 1| < δ.
8. Let f(x) = (sin 2x)/x and use a graphing utility to conjec-
ture the value of L = limx →0 f(x). Then let = 0.1 and
use the graphing utility and its trace feature to find a positive
number δ such that |f(x) − L| < if 0 < |x| < δ.
In Exercises 9–18, a positive number and the limit L of
a function f at a are given. Find a number δ such that
|f(x) − L| < if 0 < |x − a| < δ.
9. lim
x →4
2x = 8; = 0.1 10. lim
x →−2
1
2
x = −1; = 0.1
11. lim
x →−1
(7x + 5) = −2; = 0.01
12. lim
x →3
(5x − 2) = 13; = 0.01
13. lim
x →2
x2
− 4
x − 2
= 4; = 0.05
14. lim
x →−1
x2
− 1
x + 1
= −2; = 0.05
15. lim
x →4
x2
= 16; = 0.001 16. lim
x →9
√
x = 3; = 0.001
17. lim
x →5
1
x
=
1
5
; = 0.05 18. lim
x →0
|x| = 0; = 0.05
In Exercises 19–32, use Definition 2.4.1 to prove that the
stated limit is correct.
19. lim
x →5
3x = 15 20. lim
x →3
(4x − 5) = 7
21. lim
x →2
(2x − 7) = −3 22. lim
x →−1
(2 − 3x) = 5
23. lim
x →0
x2
+ x
x
= 1 24. lim
x →−3
x2
− 9
x + 3
= −6
25. lim
x →1
2x2
= 2 26. lim
x →3
(x2
− 5) = 4
27. lim
x →1/3
1
x
= 3 28. lim
x →−2
1
x + 1
= −1
29. lim
x →4
√
x = 2 30. lim
x →6
√
x + 3 = 3
31. lim
x →1
f(x) = 3, where f(x) =
x + 2, x = 1
10, x = 1
32. lim
x →2
(x2
+ 3x − 1) = 9
33. (a) Find the smallest positive number N such that for each
x in the interval (N, +ϱ), the value of the function
f(x) = 1/x2
is within 0.1 unit of L = 0.

(b) Find the smallest positive number N such that for each
x in the interval (N, +ϱ), the value of f(x) = x/(x+1)
(c) Find the largest negative number N such that for each
x in the interval (−ϱ, N), the value of the function
f(x) = 1/x3
(d) Find the largest negative number N such that for each
x in the interval (−ϱ, N), the value of the function
f(x) = x/(x + 1) is within 0.01 unit of L = 1.
34. In each part, ﬁnd the smallest positive value of N such that
for each x in the interval (N, +ϱ), the function f(x) = 1/x3
is within units of the number L = 0.
(a) = 0.1 (b) = 0.01 (c) = 0.001
(b) Find a positive number N such that
x2
1 + x2
− 1 <
for x > N.
(c) Find a negative number N such that
x2
1 + x2
− 1 <
for x < N.
x
y
1
x1 x2
Not drawn to scale
y =
x2
1 + x2
Figure Ex-35
(b) Find a positive number N such that
1
3
√
x
− 0 =
1
3
√
x
<
for x > N.
(c) Find a negative number N such that
1
3
√
x
− 0 =
1
3
√
x
<
for x < N.
x
y
y =
√x
3
1
x1
x2
Figure Ex-36
In Exercises 37–40, a positive number and the limit L of a
function f at +ϱ are given. Find a positive number N such
that |f(x) − L| < if x > N.
37. lim
x →+ϱ
1
x2
= 0; = 0.01
38. lim
x →+ϱ
1
x + 2
= 0; = 0.005
39. lim
x →+ϱ
x
x + 1
= 1; = 0.001
40. lim
x →+ϱ
4x − 1
2x + 5
= 2; = 0.1
In Exercises 41–44, a positive number and the limit L of a
function f at −ϱ are given. Find a negative number N such
that |f(x) − L| < if x < N.
41. lim
x →−ϱ
1
x + 2
= 0; = 0.005
42. lim
x →−ϱ
1
x2
= 0; = 0.01
43. lim
x →−ϱ
4x − 1
2x + 5
= 2; = 0.1
44. lim
x →−ϱ
x
x + 1
= 1; = 0.001
In Exercises 45–52, use Deﬁnition 2.4.2 or 2.4.3 to prove that
the stated limit is correct.
45. lim
x →+ϱ
1
x2
= 0 46. lim
x →−ϱ
1
x
= 0
47. lim
x →−ϱ
1
x + 2
= 0 48. lim
x →+ϱ
1
x + 2
= 0
49. lim
x →+ϱ
x
x + 1
= 1 50. lim
x →−ϱ
x
x + 1
= 1
51. lim
x →−ϱ
4x − 1
2x + 5
= 2 52. lim
x →+ϱ
4x − 1
2x + 5
= 2
53. (a) Find the largest open interval, centered at the origin on
the x-axis, such that for each x in the interval, other

2.5 Continuity 147
than the center, the values of f(x) = 1/x2
are greater
than 100.
(b) Find the largest open interval, centered at x = 1, such
that for each x in the interval, other than the center, the
values of the function
f(x) = 1/|x − 1|
are greater than 1000.
(c) Find the largest open interval, centered at x = 3, such
that for each x in the interval, other than the center, the
values of the function
f(x) = −1/(x − 3)2
are less than −1000.
(d) Find the largest open interval, centered at the origin on
the x-axis, such that for each x in the interval, other
than the center, the values of f(x) = −1/x4
are less
than −10,000.
54. In each part, find the largest open interval, centered at
x = 1, such that for each x in the interval the value of
f(x) = 1/(x − 1)2
is greater than M.
(a) M = 10 (b) M = 1000 (c) M = 100,000
In Exercises 55–60, use Definition 2.4.4 or 2.4.5 to prove that
the stated limit is correct.
55. lim
x →3
1
(x − 3)2
= +ϱ 56. lim
x →3
−1
(x − 3)2
= −ϱ
57. lim
x →0
1
|x|
= +ϱ 58. lim
x →1
1
|x − 1|
= +ϱ
59. lim
x →0
−
1
x4
= −ϱ 60. lim
x →0
1
x4
= +ϱ
In Exercises 61–66, use the remark following Definition 2.4.1
to prove that the stated limit is correct.
61. lim
x →2+
(x + 1) = 3 62. lim
x →1−
(3x + 2) = 5
63. lim
x →4+
√
x − 4 = 0 64. lim
x →0−
√
−x = 0
65. lim
x →2+
x, x > 2
3x, x ≤ 2
66. lim
x →2−
x, x > 2
3x, x ≤ 2
In Exercises 67 and 68, use the remark following Definitions
2.4.4 and 2.4.5 to prove that the stated limit is correct.
67. (a) lim
x →1+
1
1 − x
= −ϱ (b) lim
x →1−
1
1 − x
= +ϱ
68. (a) lim
x →0+
1
x
= +ϱ (b) lim
x →0−
1
x
= −ϱ
For Exercises 69 and 70, write out definitions of the four lim-
its in (18), and use your definitions to prove that the stated
limits are correct.
69. (a) lim
x →+ϱ
(x + 1) = +ϱ (b) lim
x →−ϱ
(x + 1) = −ϱ
70. (a) lim
x →+ϱ
(x2
− 3) = +ϱ (b) lim
x →−ϱ
(x3
+ 5) = −ϱ
71. Prove the result in Example 4 under the assumption that
δ ≤ 2 rather than δ ≤ 1.
72. (a) In Definition 2.4.1 there is a condition requiring that
f(x) be defined for all x in some open interval contain-
ing a, except possibly at a itself. What is the purpose
of this requirement?
(b) Why is lim
x →0
√
x = 0 an incorrect statement?
(c) Is lim
x →0.01
√
x = 0.1 a correct statement?
2.5 CONTINUITY
A moving object cannot vanish at some point and reappear someplace else to con-
tinue its motion. Thus, we perceive the path of a moving object as an unbroken curve,
without gaps, breaks, or holes. In this section, we translate “unbroken curve” into a
precise mathematical formulation called continuity, and develop some fundamental
properties of continuous curves.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
DEFINITION OF CONTINUITY
Recall from Theorem 2.2.3 that if p(x) is a polynomial and c is a real number, then
limx →c p(x) = p(c) (see Figure 2.5.1). Together with Theorem 2.2.2, we are able to
calculate limits of a variety of combinations of functions by evaluating the combination.
That is, we saw many examples of functions f (x) such that limx →c f(x) = f (c) if f(x)
is defined on an interval containing a number c. In this case, function values f(x) can be
guaranteed to be near f(c) for any x-value selected close enough to c. (See Exercise 53 for
a precise formulation of this statement.)

c
p(c)
x→c
lim p(x) = p(c)
Figure 2.5.1
On the other hand, we have also seen functions for which this nice property is not true.
For example,
f(x) =
sin(π/x), x = 0
0, x = 0
does not satisfy limx →0 f(x) = f (0), since limx →0 f(x) fails to exist.
-1 1
-1
1
x
y
Figure 2.5.2
The term continuous is used to describe the useful circumstance where the calculation
of a limit can be accomplished by mere evaluation of the function.
2.5.1 DEFINITION. A function f is said to be continuous at x = c provided the
following conditions are satisfied:
1. f(c) is defined.
2. lim
x →c
f(x) exists.
3. lim
x →c
f(x) = f(c).
If one or more of the conditions of this definition fails to hold, then we will say that f
has a discontinuity at x = c. Each function drawn in Figure 2.5.3 illustrates a discontinuity
at x = c. In Figure 2.5.3a, the function is not defined at c, violating the first condition
of Definition 2.5.1. In Figures 2.5.3b and 2.5.3c, limx →c f(x) does not exist, violating
the second condition of Definition 2.5.1. In Figure 2.5.3d, the function is defined at c and
limx →c f(x) exists, but these two values are not equal, violating the third condition of
Definition 2.5.1.
From such graphs we can develop an intuitive, geometric feel for where a function is
continuous and where it is discontinuous. Observe that continuity at c may fail due to a
“break” in the graph of the function, either due to a hole or to a jump as in Figure 2.5.3, or
perhaps due to a wild oscillation as in Figure 2.5.2. Although the intuitive interpretation of
“f is continuous at c” as “the graph of f is unbroken at c” lacks precision, it is a useful
guide in most circumstances.

2.5 Continuity 149
x
y
y = f(x)
(b)
c
x
y
y = f(x)
(c)
c
x
y
y = f(x)
(d)
c
x
y
y = f(x)
(a)
c
Figure 2.5.3
••
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
REMARK. Note that the third condition of Definition 2.5.1 really implies the first two
conditions, since it is understood in the statement limx →c f(x) = f (c) that the limit on
the left exists, the expression f(c) on the right is defined and has a finite value, and that
quantitites on the two sides are equal. Thus, when we want to establish continuity of a
function at a point our usual procedure will be to establish the validity of the third condition
only.
Example 1 Determine whether the following functions are continuous at x = 2.
f(x) =
x2
− 4
x − 2
, g(x) =



x2
− 4
x − 2
, x = 2
3, x = 2,
h(x) =



x2
− 4
x − 2
, x = 2
4, x = 2
Solution. In each case we must determine whether the limit of the function as x → 2 is
the same as the value of the function at x = 2. In all three cases the functions are identical,
except at x = 2, and hence all three have the same limit at x = 2, namely
lim
x →2
f(x) = lim
x →2
g(x) = lim
x →2
h(x) = lim
x →2
x2
− 4
x − 2
= lim
x →2
(x + 2) = 4
The function f is undefined at x = 2, and hence is not continuous at x = 2 (Figure 2.5.4a).
The function g is defined at x = 2, but its value there is g(2) = 3, which is not the same as
the limit as x approaches z; hence, g is also not continuous at x = 2 (Figure 2.5.4b). The
value of the function h at x = 2 is h(2) = 4, which is the same as the limit as x approaches
z; hence, h is continuous at x = 2 (Figure 2.5.4c). (Note that the function h could have
been written more simply as h(x) = x + 2, but we wrote it in piecewise form to emphasize
its relationship to f and g.)
2
4
x
y
y = f(x)
2
3
x
y
y = g(x)
2
4
x
y
y = h(x)
(a) (b) (c)
Figure 2.5.4
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
CONTINUITY IN APPLICATIONS
In applications, discontinuities often signal the occurrence of important physical phenom-
ena. For example, Figure 2.5.5a is a graph of voltage versus time for an underground cable
that is accidentally cut by a work crew at time t = t0 (the voltage drops to zero when the line

t
V (Voltage)
(a)
t
y (Units of inventory)
(b)
t0Line
cut Restocking occurs
y0
y1
Figure 2.5.5
is cut). Figure 2.5.5b shows the graph of inventory versus time for a company that restocks
its warehouse to y1 units when the inventory falls to y0 units. The discontinuities occur at
those times when restocking occurs.
Given the possible physical significance of discontinuities, it is important to be able to
identify discontinuities for specific functions, and to be able to make general statements
about the continuity properties of entire families of functions. This is our next goal.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
CONTINUITY ON AN INTERVAL AND
CONTINUITY OF POLYNOMIALS
If a function f is continuous at each number in an open interval (a, b), then we say that f is
continuous on (a, b). This definition applies to infinite open intervals of the form (a, +ϱ),
(−ϱ, b), and (−ϱ, +ϱ). In the case where f is continuous on (−ϱ, +ϱ), we will say that
f is continuous everywhere.
The general procedure for showing that a function is continuous everywhere is to show
that it is continuous at an arbitrary real number. For example, we showed in Theorem 2.2.3
that if p(x) is a polynomial and a is any real number, then
lim
x →a
p(x) = p(a)
Thus, we have the following result.
2.5.2 THEOREM. Polynomials are continuous everywhere.
Example 2 Show that |x| is continuous everywhere (Figure 1.2.5).
Solution. We can write |x| as
|x| =



x if x > 0
0 if x = 0
−x if x < 0
so |x| is the same as the polynomial x on the interval (0, +ϱ) and is the same as the
polynomial −x on the interval (−ϱ, 0). But polynomials are continuous everywhere, so
x = 0 is the only possible discontinuity for |x|. Since |0| = 0, to prove the continuity at
x = 0 we must show that
lim
x →0
|x| = 0 (1)
Because the formula for |x| changes at 0, it will be helpful to consider the one-sided limits
at 0 rather than the two-sided limit. We obtain
lim
x →0+
|x| = lim
x →0+
x = 0 and lim
x →0−
|x| = lim
x →0−
(−x) = 0
Thus, (1) holds and |x| is continuous at x = 0.

2.5 Continuity 151
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
SOME PROPERTIES OF
CONTINUOUS FUNCTIONS
The following theorem, which is a consequence of Theorem 2.2.2, will enable us to reach
conclusions about the continuity of functions that are obtained by adding, subtracting,
multiplying, and dividing continuous functions.
2.5.3 THEOREM. If the functions f and g are continuous at c, then
(a) f + g is continuous at c.
(b) f − g is continuous at c.
(c) fg is continuous at c.
(d) f /g is continuous at c if g(c) = 0 and has a discontinuity at c if g(c) = 0.
We will prove part (d ). The remaining proofs are similar and will be omitted.
Proof. First, consider the case where g(c) = 0. In this case f(c)/g(c) is undeﬁned, so
the function f /g has a discontinuity at c.
Next, consider the case where g(c) = 0. To prove that f /g is continuous at c, we must
show that
lim
x →c
f(x)
g(x)
=
f(c)
g(c)
(2)
Since f and g are continuous at c,
lim
x →c
f(x) = f(c) and lim
x →c
g(x) = g(c)
Thus, by Theorem 2.2.2(d )
lim
x →c
f(x)
g(x)
=
lim
x →c
f(x)
lim
x →c
g(x)
=
f(c)
g(c)
which proves (2).
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
CONTINUITY OF RATIONAL
FUNCTIONS
Since polynomials are continuous everywhere, and since rational functions are ratios of
polynomials, part (d ) of Theorem 2.5.3 yields the following result.
2.5.4 THEOREM. A rational function is continuous at every number where the de-
nominator is nonzero.
Example 3 For what values of x is there a hole or a gap in the graph of
y =
x2
− 9
x2 − 5x + 6
?
Solution. The function being graphed is a rational function, and hence is continuous at
every number where the denominator is nonzero. Solving the equation
x2
− 5x + 6 = 0
yields discontinuities at x = 2 and at x = 3.
••
•
•
•
•
•
•
•
•
•
•
•
FOR THE READER. If you use a graphing utility to generate the graph of the equation in
this example, then there is a good chance that you will see the discontinuity at x = 2 but
not at x = 3. Try it, and explain what you think is happening.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
CONTINUITY OF COMPOSITIONS
The following theorem, whose proof is given in Appendix G, will be useful for calculating
limits of compositions of functions.

2.5.5 THEOREM. If limx →c g(x) = L and if the function f is continuous at L, then
limx →c f(g(x)) = f(L). That is,
lim
x →c
f(g(x)) = f lim
x →c
g(x)
This equality remains valid if limx →c is replaced everywhere by one of limx →c+ ,
limx →c− , limx →+ϱ, or limx →−ϱ.
In words, this theorem states:
A limit symbol can be moved through a function sign provided the limit of the expres-
sion inside the function sign exists and the function is continuous at this limit.
Example 4 We know from Example 2 that the function |x| is continuous everywhere;
thus, it follows that if limx →a g(x) exists, then
lim
x →a
|g(x)| = lim
x →a
g(x) (3)
That is, a limit symbol can be moved through an absolute value sign, provided the limit of
the expression inside the absolute value signs exists. For example,
lim
x →3
|5 − x2
| = lim
x →3
(5 − x2
) = | − 4| = 4
The following theorem is concerned with the continuity of compositions of functions;
the ﬁrst part deals with continuity at a speciﬁc number, and the second part with continuity
everywhere.
2.5.6 THEOREM.
(a) If the function g is continuous at c, and the function f is continuous at g(c), then
the composition f ◦g is continuous at c.
(b) If the function g is continuous everywhere and the function f is continuous every-
where, then the composition f ◦g is continuous everywhere.
Proof. We will prove part (a) only; the proof of part (b) can be obtained by applying part
(a) at an arbitrary number c. To prove that f ◦g is continuous at c, we must show that the
value of f ◦g and the value of its limit are the same at x = c. But this is so, since we can
write
lim
x →c
(f ◦g)(x) = lim
x →c
f(g(x)) = f( lim
x →c
g(x)) = f(g(c)) = (f ◦g)(c)
Theorem 2.5.5 g is continuous at c.
We know from Example 2 that the function |x| is continuous everywhere. Thus, if g(x)
is continuous at c, then by part (a) of Theorem 2.5.6, the function |g(x)| must also be
continuous at c; and, more generally, if g(x) is continuous everywhere, then so is |g(x)|.
Stated informally:
The absolute value of a continuous function is continuous.
For example, the polynomial g(x) = 4 − x2
is continuous everywhere, so we can conclude
that the function |4 − x2
| is also continuous everywhere (Figure 2.5.6).
-4 -3 -2 -1 1 2 3 4
1
2
3
4
5
x
y y = |4 – x2|
Figure 2.5.6
••
•
•
•
•
•
•
FOR THE READER. Can the absolute value of a function that is not continuous be contin-
uous? Justify your answer.

2.5 Continuity 153
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
CONTINUITY FROM THE LEFT
AND RIGHT
Because Definition 2.5.1 involves a two-sided limit, that definition does not generally apply
at the endpoints of a closed interval [a, b] or at the endpoint of an interval of the form
[a, b), (a, b], (−ϱ, b], or [a, +ϱ). To remedy this problem, we will agree that a function
is continuous at an endpoint of an interval if its value at the endpoint is equal to the appro-
priate one-sided limit at that endpoint. For example, the function graphed in Figure 2.5.7 is
continuous at the right endpoint of the interval [a, b] because
lim
x →b−
f(x) = f(b)
but it is not continuous at the left endpoint because
lim
x →a+
f(x) = f(a)
In general, we will say a function f is continuous from the left at c if
lim
x →c−
f(x) = f(c)
and is continuous from the right at c if
lim
x →c+
f(x) = f(c)
Using this terminology we define continuity on a closed interval as follows.
x
y
y = f(x)
a b
Figure 2.5.7
2.5.7 DEFINITION. A function f is said to be continuous on a closed interval [a, b]
if the following conditions are satisfied:
1. f is continuous on (a, b).
2. f is continuous from the right at a.
3. f is continuous from the left at b.
••
•
•
•
•
•
•
FOR THE READER. We leave it for you to modify this definition appropriately so that it
applies to intervals of the form [a, +ϱ), (−ϱ, b], (a, b], and [a, b).
Example 5 What can you say about the continuity of the function f(x) = 9 − x2?
Solution. Because the natural domain of this function is the closed interval [−3, 3], we
will need to investigate the continuity of f on the open interval (−3, 3) and at the two
endpoints. If c is any number in the interval (−3, 3), then it follows from Theorem 2.2.2(e)
that
lim
x →c
f(x) = lim
x →c
9 − x2 = lim
x →c
(9 − x2) = 9 − c2 = f(c)
which proves f is continuous at each number in the interval (−3, 3). The function f is also
continuous at the endpoints since
lim
x →3−
f(x) = lim
x →3−
9 − x2 = lim
x →3−
(9 − x2) = 0 = f(3)
lim
x →−3+
f(x) = lim
x →−3+
9 − x2 = lim
x →−3+
(9 − x2) = 0 = f(−3)
Thus, f is continuous on the closed interval [−3, 3].
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
THE INTERMEDIATE-VALUE
THEOREM
Figure 2.5.8 shows the graph of a function that is continuous on the closed interval [a, b].
The figure suggests that if we draw any horizontal line y = k, where k is between f(a)
and f(b), then that line will cross the curve y = f(x) at least once over the interval [a, b].
Stated in numerical terms, if f is continuous on [a, b], then the function f must take on
every value k between f(a) and f(b) at least once as x varies from a to b. For example,
the polynomial p(x) = x5
− x + 3 has a value of 3 at x = 1 and a value of 33 at x = 2.
Thus, it follows from the continuity of p that the equation x5
− x + 3 = k has at least one

solution in the interval [1, 2] for every value of k between 3 and 33. This idea is stated more
precisely in the following theorem.
2.5.8 THEOREM (Intermediate-Value Theorem). If f is continuous on a closed interval
[a, b] and k is any number between f(a) and f(b), inclusive, then there is at least one
number x in the interval [a, b] such that f(x) = k.
x
y
f(a)
k
f(b)
a bx
Figure 2.5.8
Although this theorem is intuitively obvious, its proof depends on a mathematically precise
development of the real number system, which is beyond the scope of this text.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
APPROXIMATING ROOTS USING
THE INTERMEDIATE-VALUE
THEOREM
A variety of problems can be reduced to solving an equation f(x) = 0 for its roots. Some-
times it is possible to solve for the roots exactly using algebra, but often this is not possible
and one must settle for decimal approximations of the roots. One procedure for approxi-
mating roots is based on the following consequence of the Intermediate-Value Theorem.
2.5.9 THEOREM. If f is continuous on [a, b], and if f(a) and f(b) are nonzero and
have opposite signs, then there is at least one solution of the equation f(x) = 0 in the
interval (a, b).
This result, which is illustrated in Figure 2.5.9, can be proved as follows.
x
y
f(a) > 0
f(b) < 0
f(x) = 0
a
b
Figure 2.5.9
Proof. Since f(a) and f(b) have opposite signs, 0 is between f(a) and f(b). Thus, by
the Intermediate-Value Theorem there is at least one number x in the interval [a, b] such
that f(x) = 0. However, f(a) and f(b) are nonzero, so x must lie in the interval (a, b),
which completes the proof.
Before we illustrate how this theorem can be used to approximate roots, it will be helpful
to discuss some standard terminology for describing errors in approximations. If x is an
approximation to a quantity x0, then we call
= |x − x0|
the absolute error or (less precisely) the error in the approximation. The terminology in
Table 2.5.1 is used to describe the size of such errors:
Table 2.5.1
error description
|x – x0| ≤ 0.1
|x – x0| ≤ 0.01
|x – x0| ≤ 0.001
|x – x0| ≤ 0.0001
|x – x0| ≤ 0.5
|x – x0| ≤ 0.05
|x – x0| ≤ 0.005
|x – x0| ≤ 0.0005
x approximates x0 with an error of at most 0.1.
x approximates x0 to the nearest integer.
x approximates x0 to 1 decimal place (i.e., to the nearest tenth).
x approximates x0 to 2 decimal places (i.e., to the nearest hundredth).
x approximates x0 to 3 decimal places (i.e., to the nearest thousandth).
Example 6 The equation
x3
− x − 1 = 0
cannot be solved algebraically very easily because the left side has no simple factors.
However, if we graph p(x) = x3
− x − 1 with a graphing utility (Figure 2.5.10), then we
are led to conjecture that there is one real root and that this root lies inside the interval [1, 2].

2.5 Continuity 155
The existence of a root in this interval is also conﬁrmed by Theorem 2.5.9, since p(1) = −1
and p(2) = 5 have opposite signs. Approximate this root to two decimal-place accuracy.
x
y
y = x3
– x – 1
2
2
Figure 2.5.10
Solution. Our objective is to approximate the unknown root x0 with an error of at most
0.005. It follows that if we can ﬁnd an interval of length 0.01 that contains the root, then the
midpoint of that interval will approximate the root with an error of at most 0.01/2 = 0.005,
which will achieve the desired accuracy.
We know that the root x0 lies in the interval [1, 2]. However, this interval has length
1, which is too large. We can pinpoint the location of the root more precisely by dividing
the interval [1, 2] into 10 equal parts and evaluating p at the points of subdivision using
a calculating utility (Table 2.5.2). In this table p(1.3) and p(1.4) have opposite signs, so
we know that the root lies in the interval [1.3, 1.4]. This interval has length 0.1, which is
still too large, so we repeat the process by dividing the interval [1.3, 1.4] into 10 parts and
evaluating p at the points of subdivision; this yields Table 2.5.3, which tells us that the root
is inside the interval [1.32, 1.33] (Figure 2.5.11). Since this interval has length 0.01, its
midpoint 1.325 will approximate the root with an error of at most 0.005. Thus, x0 ≈ 1.325
to two decimal-place accuracy.
Table 2.5.2
1
–1
1.1
–0.77
1.2
–0.47
1.3
–0.10 0.34
1.5
0.88
1.6
1.50
1.7
2.21
1.8
3.03
1.4x
f(x)
1.9
3.96
2
5
Table 2.5.3
1.3
–0.103
1.31
–0.062
1.32
–0.020
1.33
0.023 0.066
1.35
0.110
1.36
0.155
1.37
0.201
1.38
0.248
1.34x
f(x)
1.39
0.296
1.4
0.344
1.322 1.324 1.326 1.328 1.330
-0.02
-0.01
0.01
0.02
x
y
y = p(x) = x3
– x – 1
Figure 2.5.11
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
APPROXIMATING ROOTS BY
ZOOMING WITH A GRAPHING
UTILITY
The method illustrated in Example 6 can also be implemented with a graphing utility as
follows.
Step 1. Figure 2.5.12a shows the graph of f in the window [−5, 5]×[−5, 5]
with xScl = 1 and yScl = 1. That graph places the root between
x = 1 and x = 2.
Step 2. Since we know that the root lies between x = 1 and x = 2, we will
zoom in by regraphing f over an x-interval that extends between
these values and in which xScl = 0.1. The y-interval and yScl are not
critical, as long as the y-interval extends above and below the x-axis.
Figure 2.5.12b shows the graph of f in the window [1, 2] × [−1, 1]
with xScl = 0.1 and yScl = 0.1. That graph places the root between
x = 1.3 and x = 1.4.

Step 3. Since we know that the root lies between x = 1.3 and x = 1.4, we
will zoom in again by regraphing f over an x-interval that extends
betweenthesevaluesandinwhichxScl = 0.01.Figure2.5.12cshows
the graph of f in the window [1.3, 1.4] × [−0.1, 0.1] with xScl =
0.01 and yScl = 0.01. That graph places the root between x = 1.32
and x = 1.33.
Step 4. Since the interval in Step 3 has length 0.01, its midpoint 1.325 ap-
proximates the root with an error of at most 0.005, so x0 ≈ 1.325 to
two decimal-place accuracy.
Figure 2.5.12
[–5, 5] × [–5, 5]
xScl = 1, yScl = 1
[1, 2] × [–1, 1]
xScl = 0.1, yScl = 0.1
[1.3, 1.4] × [–0.1, 0.1]
xScl = 0.01, yScl = 0.01
(b) (c)(a)
••
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
REMARK. To say that x approximates x0 to n decimal places does not mean that the first
n decimal places of x and x0 will be the same when the numbers are rounded to n decimal
places. For example, x = 1.084 approximates x0 = 1.087 to two decimal places because
|x − x0| = 0.003(<0.005). However, if we round these values to two decimal places, then
we obtain x ≈ 1.08 and x0 ≈ 1.09. Thus, if you approximate a number to n decimal places,
then you should display that approximation to at least n + 1 decimal places to preserve the
accuracy.
••
•
•
•
•
•
•
FOR THE READER. Use a graphing or calculating utility to show that the root x0 in Example
6 can be approximated as x0 ≈ 1.3245 to three decimal-place accuracy.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
In Exercises 1–4, let f be the function whose graph is shown.
On which of the following intervals, if any, is f continuous?
(a) [1, 3] (b) (1, 3) (c) [1, 2]
(d) (1, 2) (e) [2, 3] (f) (2, 3)
Foreachintervalonwhichf isnotcontinuous,indicatewhich
conditions for the continuity of f do not hold.
1.
1 2 3
x
y
2.
1 2 3
x
y
3.
1 2 3
x
y 4.
1 2 3
x
y
In Exercises 5 and 6, find all values of c such that the specified
function has a discontinuity at x = c. For each such value of
c, determine which conditions of Definition 2.5.1 fail to be
satisfied.

2.5 Continuity 157
5. (a) The function f in Exercise 1 of Section 2.1.
(b) The function F in Exercise 5 of Section 2.1.
(c) The function f in Exercise 9 of Section 2.1.
6. (a) The function f in Exercise 2 of Section 2.1.
(b) The function F in Exercise 6 of Section 2.1.
(c) The function f in Exercise 10 of Section 2.1.
7. Suppose that f and g are continuous functions such that
f(2) = 1 and lim
x →2
[f(x) + 4g(x)] = 13. Find
(a) g(2) (b) lim
x →2
g(x).
8. Suppose that f and g are continuous functions such that
lim
x →3
g(x) = 5 and f(3) = −2. Find lim
x →3
[f(x)/g(x)].
9. In each part sketch the graph of a function f that satisfies
the stated conditions.
(a) f is continuous everywhere except at x = 3, at which
point it is continuous from the right.
(b) f has a two-sided limit at x = 3, but it is not continuous
at x = 3.
(c) f is not continuous at x = 3, but if its value at x = 3
is changed from f(3) = 1 to f(3) = 0, it becomes
continuous at x = 3.
(d) f is continuous on the interval [0, 3) and is defined on
the closed interval [0, 3]; but f is not continuous on the
interval [0, 3].
10. Find formulas for some functions that are continuous on the
intervals (−ϱ, 0) and (0, +ϱ), but are not continuous on the
interval (−ϱ, +ϱ).
11. A student parking lot at a university charges $2.00 for the
first half hour (or any part) and $1.00 for each subsequent
half hour (or any part) up to a daily maximum of $10.00.
(a) Sketch a graph of cost as a function of the time parked.
(b) Discuss the significance of the discontinuities in the
graph to a student who parks there.
12. In each part determine whether the function is continuous
or not, and explain your reasoning.
(a) The Earth’s population as a function of time
(b) Your exact height as a function of time
(c) The cost of a taxi ride in your city as a function of the
distance traveled
(d) The volume of a melting ice cube as a function of time
In Exercises 13–24, find the values of x (if any) at which f
is not continuous.
13. f(x) = x3
− 2x + 3 14. f(x) = (x − 5)17
15. f(x) =
x
x2 + 1
16. f(x) =
x
x2 − 1
17. f(x) =
x − 4
x2 − 16
18. f(x) =
3x + 1
x2 + 7x − 2
19. f(x) =
x
|x| − 3
20. f(x) =
5
x
+
2x
x + 4
21. f(x) = |x3
− 2x2
| 22. f(x) =
x + 3
|x2 + 3x|
23. f(x) =



2x + 3, x ≤ 4
7 +
16
x
, x > 4
24. f(x) =



3
x − 1
, x = 1
3, x = 1
25. Find a value for the constant k, if possible, that will make
the function continuous everywhere.
(a) f(x) =
7x − 2, x ≤ 1
kx2
, x > 1
(b) f(x) =
kx2
, x ≤ 2
2x + k, x > 2
26. On which of the following intervals is
f(x) =
1
√
x − 2
continuous?
(a) [2, +ϱ) (b) (−ϱ, +ϱ) (c) (2, +ϱ) (d) [1, 2)
A function f is said to have a removable discontinuity at
x = c if limx →c f(x) exists but f is not continuous at x = c,
either because f is not defined at c or because the definition
for f(c) differs from the value of the limit. This terminology
will be needed in Exercises 27–30.
27. (a) Sketch the graph of a function with a removable dis-
continuity at x = c for which f(c) is undefined.
(b) Sketch the graph of a function with a removable dis-
continuity at x = c for which f(c) is defined.
28. (a) The terminology removable discontinuity is appropri-
ate because a removable discontinuity of a function f
at x = c can be “removed” by redefining the value of
f appropriately at x = c. What value for f(c) removes
the discontinuity?
(b) Show that the following functions have removable dis-
continuities at x = 1, and sketch their graphs.
f(x) =
x2
− 1
x − 1
and g(x) =



1, x > 1
0, x = 1
1, x < 1
(c) What values should be assigned to f(1) and g(1) to
remove the discontinuities?
In Exercises 29 and 30, find the values of x (if any) at which
f is not continuous, and determine whether each such value
is a removable discontinuity.
29. (a) f(x) =
|x|
x
(b) f(x) =
x2
+ 3x
x + 3
(c) f(x) =
x − 2
|x| − 2

30. (a) f(x) =
x2
− 4
x3 − 8
(b) f(x) =
2x − 3, x ≤ 2
x2
, x > 2
(c) f(x) =
3x2
+ 5, x = 1
6, x = 1
31. (a) Use a graphing utility to generate the graph of the func-
tion f(x) = (x + 3)/(2x2
+ 5x − 3), and then use
the graph to make a conjecture about the number and
locations of all discontinuities.
(b) Check your conjecture by factoring the denominator.
32. (a) Use a graphing utility to generate the graph of the func-
tion f(x) = x/(x3
− x + 2), and then use the graph to
make a conjecture about the number and locations of
all discontinuities.
(b) Use the Intermediate-Value Theorem to approximate
the location of all discontinuities to two decimal places.
33. Prove that f(x) = x3/5
is continuous everywhere, carefully
justifying each step.
34. Prove that f(x) = 1/ x4 + 7x2 + 1 is continuous every-
where, carefully justifying each step.
35. Let f and g be discontinuous at c. Give examples to show
that
(a) f + g can be continuous or discontinuous at c
(b) fg can be continuous or discontinuous at c.
36. Prove Theorem 2.5.4.
37. Prove:
(a) part (a) of Theorem 2.5.3
(b) part (b) of Theorem 2.5.3
(c) part (c) of Theorem 2.5.3.
38. Prove: If f and g are continuous on [a, b], and f(a) > g(a),
f(b) < g(b), then there is at least one solution of the equa-
tion f(x) = g(x) in (a, b). [Hint: Consider f(x) − g(x).]
39. Give an example of a function f that is defined on a closed
interval, and whose values at the endpoints have opposite
signs, but for which the equation f(x) = 0 has no solution
in the interval.
40. Use the Intermediate-Value Theorem to show that there is a
square with a diagonal length that is between r and 2r and
an area that is half the area of a circle of radius r.
41. Use the Intermediate-Value Theorem to show that there is
a right circular cylinder of height h and radius less than r
whose volume is equal to that of a right circular cone of
height h and radius r.
In Exercises 42 and 43, show that the equation has at least
one solution in the given interval.
42. x3
− 4x + 1 = 0; [1, 2] 43. x3
+x2
−2x = 1; [−1, 1]
44. Prove: If p(x) is a polynomial of odd degree, then the equa-
tion p(x) = 0 has at least one real solution.
45. The accompanying figure shows the graph of y = x4
+x−1.
Use the method of Example 6 to approximate the x-
intercepts with an error of at most 0.05.
[–5, 4] × [–3, 6]
xScl = 1, yScl = 1
Figure Ex-45
46. Use a graphing utility to solve the problem in Exercise 45
by zooming.
47. The accompanying figure shows the graph of y = 5−x−x4
.
Use the method of Example 6 to approximate the roots of
the equation 5−x −x4
= 0 to two decimal-place accuracy.
[–5, 4] × [–3, 6]
xScl = 1, yScl = 1
Figure Ex-47
48. Use a graphing utility to solve the problem in Exercise 47
by zooming.
49. Use the fact that
√
5 is a solution of x2
− 5 = 0 to approxi-
mate
√
5 with an error of at most 0.005.
50. Prove that if a and b are positive, then the equation
a
x − 1
+
b
x − 3
= 0
has at least one solution in the interval (1, 3).
51. A sphere of unknown radius x consists of a spherical core
and a coating that is 1 cm thick (see the accompanying fig-
ure). Given that the volume of the coating and the volume of
the core are the same, approximate the radius of the sphere
to three decimal-place accuracy.
1 cm
x
Figure Ex-51

2.6 Limits and Continuity of Trigonometric Functions 159
52. A monk begins walking up a mountain road at 12:00 noon
and reaches the top at 12:00 midnight. He meditates and
rests until 12:00 noon the next day, at which time he begins
walking down the same road, reaching the bottom at 12:00
midnight. Show that there is at least one point on the road
that he reaches at the same time of day on the way up as on
the way down.
53. Let f be defined at c. Prove that f is continuous at c if, given
> 0, there exists a δ > 0 such that |f(x) − f(c)| < if
|x − c| < δ.
2.6 LIMITS AND CONTINUITY OF TRIGONOMETRIC FUNCTIONS
In this section we will investigate the continuity properties of the trigonometric func-
tions, and we will discuss some important limits involving these functions.
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
CONTINUITY OF TRIGONOMETRIC
FUNCTIONS
Before we begin, recall that in the expressions sin x, cos x, tan x, cot x, sec x, and csc x it
is understood that x is in radian measure.
In trigonometry, the graphs of sin x and cos x are drawn as continuous curves (Fig-
ure 2.6.1). To actually prove that these functions are continuous everywhere, we must show
that the following equalities hold for every real number c:
lim
x →c
sin x = sin c and lim
x →c
cos x = cos c (1–2)
Although we will not formally prove these results, we can make them plausible by consid-
ering the behavior of the point P(cos x, sin x) as it moves around the unit circle. For this
purpose, view c as a fixed angle in radian measure, and let Q(cos c, sin c) be the correspond-
ing point on the unit circle. As x →c (i.e., as the angle x approaches the angle c), the point
P moves along the circle toward Q, and this implies that the coordinates of P approach the
corresponding coordinates of Q; that is, cos x →cos c, and sin x →sin c (Figure 2.6.2).
-1
1
y = sin x
cCO o
x
y
-1
1
y = cos x
cCO o
x
y
Figure 2.6.1
Formulas (1) and (2) can be used to find limits of the remaining trigonometric functions
by expressing them in terms of sin x and cos x; for example, if cos c = 0, then
lim
x →c
tan x = lim
x →c
sin x
cos x
=
sin c
cos c
= tan c
Thus, we are led to the following theorem.
2.6.1 THEOREM. If c is any number in the natural domain of the stated trigonometric
function, then
lim
x →c
sin x = sin c lim
x →c
cos x = cos c lim
x →c
tan x = tan c
lim
x →c
csc x = csc c lim
x →c
sec x = sec c lim
x →c
cot x = cot c
Q(cos c, sin c)
P(cos x, sin x)
x
c
Figure 2.6.2
It follows from this theorem, for example, that sin x and cos x are continuous everywhere
and that tan x is continuous, except at the points where it is undefined.

Example 1 Find the limit
lim
x →1
cos
x2
− 1
x − 1
Solution. Recall from the last section that since the cosine function is continuous every-
where,
lim
x →1
cos(g(x)) = cos( lim
x →1
g(x))
provided limx →1 g(x) exists. Thus,
lim
x →1
cos
x2
− 1
x − 1
= lim
x →1
cos(x + 1) = cos lim
x →1
(x + 1) = cos 2
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
OBTAINING LIMITS BY SQUEEZING
In Section 2.1 we used the numerical evidence in Table ?? to conjecture that
lim
x →0
sin x
x
= 1 (3)
However, it is not a simple matter to establish this limit with certainty. The difficulty is
that the numerator and denominator both approach zero as x →0. As discussed in Section
2.2, such limits are called indeterminate forms of type 0/0. Sometimes indeterminate forms
of this type can be established by manipulating the ratio algebraically, but in this case no
simple algebraic manipulation will work, so we must look for other methods.
The problem with indeterminate forms of type 0/0 is that there are two conflicting
influences at work: as the numerator approaches 0 it drives the magnitude of the ratio
toward 0, and as the denominator approaches 0 it drives the magnitude of the ratio toward ±ϱ
(depending on the sign of the expression). The limiting behavior of the ratio is determined
by the precise way in which these influences offset each other. Later in this text we will
discuss general methods for attacking indeterminate forms, but for the limit in (3) we can
use a method called squeezing.
In the method of squeezing one proves that a function f has a limit L at a number c by
trapping the function between two other functions, g and h, whose limits at c are known to be
L(Figure2.6.3).Thisistheideabehindthefollowingtheorem,whichwestatewithoutproof.
x
y
c
L
y = h(x)
y = g(x)
y = f(x)
Figure 2.6.3
O o
1 x
y
O o
1 x
y
x→0
lim = 1
sin x
x
x→0
lim = 0
1 – cos x
x
y =
sin x
x
y =
1 – cos x
x
Figure 2.6.4
2.6.2 THEOREM (The Squeezing Theorem). Let f, g, and h be functions satisfying
g(x) ≤ f(x) ≤ h(x)
for all x in some open interval containing the number c, with the possible exception that
the inequalities need not hold at c. If g and h have the same limit as x approaches c, say
lim
x →c
g(x) = lim
x →c
h(x) = L
then f also has this limit as x approaches c, that is,
lim
x →c
f(x) = L
••
•
•
•
•
•
•
•
FOR THE READER. The Squeezing Theorem also holds for one-sided limits and limits at
+ϱ and −ϱ. How do you think the hypotheses of the theorem would change in those cases?
The usefulness of the Squeezing Theorem will be evident in our proof of the following
theorem (Figure 2.6.4).
2.6.3 THEOREM.
(a) lim
x →0
sin x
x
= 1 (b) lim
x →0
1 − cos x
x
= 0

However, before giving the proof, it will be helpful to review the formula for the area A of
a sector with radius r and a central angle of θ radians (Figure 2.6.5). The area of the sector
can be derived by setting up the following proportion to the area of the entire circle:
A
πr2
=
θ
2π
area of the sector
area of the circle
=
central angle of the sector
central angle of the circle
From this we obtain the formula
A = 1
2
r2
θ (4)
Now we are ready for the proof of Theorem 2.6.3.
u
r
Area = A
Figure 2.6.5
Proof (a). In this proof we will interpret x as an angle in radian measure, and we will
assume to start that 0 < x < π/2. It follows from Formula (4) that the area of a sector of
radius 1 and central angle x is x/2. Moreover, it is suggested by Figure 2.6.6 that the area
of this sector lies between the areas of two triangles, one with area (tan x)/2 and one with
area (sin x)/2. Thus,
tan x
2
≥
x
2
≥
sin x
2
Multiplying through by 2/(sin x) yields
1
cos x
≥
x
sin x
≥ 1
and then taking reciprocals and reversing the inequalities yields
cos x ≤
sin x
x
≤ 1 (5)
Moreover, these inequalities also hold for −π/2 < x < 0, since replacing x by −x in (5)
and using the identities sin(−x) = − sin x and cos(−x) = cos x leaves the inequalities
unchanged (verify). Finally, since the functions cos x and 1 both have limits of 1 as x →0,
it follows from the Squeezing Theorem that (sin x)/x also has a limit of 1 as x →0.
Figure 2.6.6
1
1
x (1, 0)
(1, tan x)
tan x
(cos x, sin x)
x
1
x
1
x
Area of triangle Area of sector Area of triangle≥
≥ ≥
≥
tan x
2
sin x
2
x
2
sin x
Proof (b). For this proof we will use the limit in part (a), the continuity of the sine function,
and the trigonometric identity sin2
x = 1 − cos2
x. We obtain
lim
x →0
1 − cos x
x
= lim
x →0
1 − cos x
x
·
1 + cos x
1 + cos x
= lim
x →0
sin2
x
(1 + cos x)x
= lim
x →0
sin x
x
lim
x →0
sin x
1 + cos x
= (1)
0
1 + 1
= 0
Example 2 Find
(a) lim
x →0
tan x
x
(b) lim
θ →0
sin 2θ
θ
(c) lim
x →0
sin 3x
sin 5x

Solution (a).
lim
x →0
tan x
x
= lim
x →0
sin x
x
·
1
cos x
= (1)(1) = 1
Solution (b). The trick is to multiply and divide by 2, which will make the denominator
the same as the argument of the sine function [just as in Theorem 2.6.3(a)]:
lim
θ →0
sin 2θ
θ
= lim
θ →0
2 ·
sin 2θ
2θ
= 2 lim
θ →0
sin 2θ
2θ
Now make the substitution x = 2θ, and use the fact that x →0 as θ →0. This yields
lim
θ →0
sin 2θ
θ
= 2 lim
θ →0
sin 2θ
2θ
= 2 lim
x →0
sin x
x
= 2(1) = 2
Solution (c).
lim
x →0
sin 3x
sin 5x
= lim
x →0
sin 3x
x
sin 5x
x
= lim
x →0
3 ·
sin 3x
3x
5 ·
sin 5x
5x
=
3 · 1
5 · 1
=
3
5
••
•
•
•
•
•
•
•
FOR THE READER. Use a graphing utility to confirm the limits in the last example graph-
ically, and if you have a CAS, then use it to obtain the limits.
Example 3 Make conjectures about the limits
(a) lim
x →0
sin
1
x
(b) lim
x →0
x sin
1
x
and confirm your conclusions by generating the graphs of the functions near x = 0 using a
graphing utility.
Solution (a). Since 1/x → +ϱ as x → 0+
, we can view sin(1/x) as the sine of an angle
that increases indefinitely as x →0+
. As this angle increases, the function sin(1/x) keeps
oscillating between −1 and 1 without approaching a limit. Similarly, there is no limit from
the left since 1/x → −ϱ as x → 0−
. These conclusions are consistent with the graph of
y = sin(1/x) shown in Figure 2.6.7a. Observe that the oscillations become more and more
rapid as x approaches 0 because 1/x increases (or decreases) more and more rapidly as x
approaches 0.
Solution (b). If x > 0, −x ≤ x sin(1/x) ≤ x, and if x < 0, x ≤ x sin(1/x) ≤ −x.
Thus, for x = 0, −|x| ≤ x sin(1/x) ≤ |x|. Since both |x| → 0 and −|x| → 0 as x → 0,
the Squeezing Theorem applies and we can conclude that x sin(1/x)→0 as x →0. This is
illustrated in Figure 2.6.7b.
-1 1
-1
1
x
y
y = sin ( )x
1
(a)
y = x sin ( )x
1
x
y
y = |x|
y = –|x|
(b)
Figure 2.6.7
••
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
REMARK. It follows from part (b) of this example that the function
f(x) =
x sin(1/x), x = 0
0, x = 0
is continuous at x = 0, since the value of the function and the value of the limit are the
same at 0. This shows that the behavior of a function can be very complex in the vicinity
of an x-value c, even though the function is continuous at c.

EXERCISE SET 2.6 Graphing Calculator C CAS
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
In Exercises 1–10, ﬁnd the discontinuities, if any.
1. f(x) = sin(x2
− 2) 2. f(x) = cos
x
x − π
3. f(x) = cot x 4. f(x) = sec x
5. f(x) = csc x 6. f(x) =
1
1 + sin2
x
7. f(x) = | cos x| 8. f(x) = 2 + tan2 x
9. f(x) =
1
1 − 2 sin x
10. f(x) =
3
5 + 2 cos x
11. Use Theorem 2.5.6 to show that the following functions
are continuous everywhere by expressing them as compo-
sitions of simpler functions that are known to be continuous.
(a) sin(x3
+ 7x + 1) (b) |sin x|
(c) cos3
(x + 1) (d)
√
3 + sin 2x
(e) sin(sin x) (f) cos5
x − 2 cos3
x + 1
12. (a) Prove that if g(x) is continuous everywhere, then so are
sin(g(x)), cos(g(x)), g(sin(x)), and g(cos(x)).
(b) Illustrate the result in part (a) with some of your own
choices for g.
Find the limits in Exercises 13–35.
13. lim
x →+ϱ
cos
1
x
14. lim
x →+ϱ
sin
2
x
15. lim
x →+ϱ
sin
πx
2 − 3x
16. lim
h→0
sin h
2h
17. lim
θ →0
sin 3θ
θ
18. lim
θ →0+
sin θ
θ2
19. lim
x →0−
sin x
|x|
20. lim
x →0
sin2
x
3x2
21. lim
x →0+
sin x
5
√
x
22. lim
x →0
sin 6x
sin 8x
23. lim
x →0
tan 7x
sin 3x
24. lim
θ →0
sin2
θ
θ
25. lim
h→0
h
tan h
26. lim
h→0
sin h
1 − cos h
27. lim
θ →0
θ2
1 − cos θ
28. lim
x →0
x
cos 1
2
π − x
29. lim
θ →0
θ
cos θ
30. lim
t →0
t2
1 − cos2 t
31. lim
h→0
1 − cos 5h
cos 7h − 1
32. lim
x →0+
sin
1
x
33. lim
x →0+
cos
1
x
34. lim
x →0
x2
− 3 sin x
x
35. lim
x →0
2x + sin x
x
In Exercises 36–39: (i) Construct a table to estimate the limit
by evaluating the function near the limiting value. (ii) Find
the exact value of the limit.
36. lim
x →5
sin(x − 5)
x2 − 25
37. lim
x →2
sin(2x − 4)
x2 − 4
38. lim
x →−2
sin(x2
+ 3x + 2)
x + 2
39. lim
x →−1
sin(x2
+ 3x + 2)
x3 + 1
40. Find a value for the constant k that makes
f(x) =



sin 3x
x
, x = 0
k, x = 0
41. Find a nonzero value for the constant k that makes
f(x) =



tan kx
x
, x < 0
3x + 2k2
, x ≥ 0
42. Is
f(x) =



sin x
|x|
, x = 0
1, x = 0
continuous at x = 0?
43. In each part, ﬁnd the limit by making the indicated substi-
tution.
(a) lim
x →+ϱ
x sin
1
x
; t =
1
x
(b) lim
x →−ϱ
x 1 − cos
1
x
; t =
1
x
(c) lim
x →π
π − x
sin x
. [Hint: Let t = π − x.]
44. Find lim
x →2
cos(π/x)
x − 2
; t =
π
2
−
π
x
.
45. Find lim
x →1
sin(πx)
x − 1
. 46. Find lim
x →π/4
tan x−1
x−π/4
.
47. Use the Squeezing Theorem to show that
lim
x →0
x cos
50π
x
= 0
and illustrate the principle involved by using a graphing util-
ity to graph y = |x|, y = −|x|, and y = x cos(50π/x) on
the same screen in the window [−1, 1] × [−1, 1].
48. Use the Squeezing Theorem to show that
lim
x →0
x2
sin
50π
3
√
x
= 0

and illustrate the principle involved by using a graphing util-
ity to graph y = x2
, y = −x2
, and y = x2
sin(50π/ 3
√
x ) on
the same screen in the window [−0.5, 0.5]×[−0.25, 0.25].
49. Sketch the graphs of y = 1 − x2
, y = cos x, and y = f(x),
where f is a function that satisfies the inequalities
1 − x2
≤ f(x) ≤ cos x
for all x in the interval (−π/2, π/2). What can you say about
the limit of f(x) as x →0? Explain your reasoning.
50. Sketch the graphs of y = 1/x, y = −1/x, and y = f(x),
where f is a function that satisfies the inequalities
−
1
x
≤ f(x) ≤
1
x
for all x in the interval [1, +ϱ). What can you say about the
limit of f(x) as x →+ϱ? Explain your reasoning.
51. Find formulas for functions g and h such that g(x)→0 and
h(x)→0 as x →+ϱ and such that
g(x) ≤
sin x
x
≤ h(x)
for positive values of x. What can you say about the limit
lim
x →+ϱ
sin x
x
?
Explain your reasoning.
52. Draw pictures analogous to Figure 2.6.3 that illustrate the
Squeezing Theorem for limits of the forms limx →+ϱ f(x)
and limx →−ϱ f(x).
Recall that unless stated otherwise the variable x in trigono-
metric functions such as sin x and cos x is assumed to be in
radian measure. The limits in Theorem 2.6.3 are based on
that assumption. Exercises 53 and 54 explore what happens
to those limits if degree measure is used for x.
53. (a) Show that if x is in degrees, then
lim
x →0
sin x
x
=
π
180
(b) Confirm that the limit in part (a) is consistent with the
results produced by your calculating utility by setting
the utility to degree measure and calculating (sin x)/x
for some values of x that get closer and closer to 0.
54. What is the limit of (1−cos x)/x as x →0 if x is in degrees?
55. It follows from part (a) of Theorem 2.6.3 that if θ is small
(near zero) and measured in radians, then one should expect
the approximation
sin θ ≈ θ
to be good.
(a) Find sin 10◦
using a calculating utility.
(b) Estimate sin 10◦
using the approximation above.
56. (a) Use the approximation of sin θ that is given in Exer-
cise 55 together with the identity cos 2α = 1 − 2 sin2
α
with α = θ/2 to show that if θ is small (near zero)
and measured in radians, then one should expect the
approximation
cos θ ≈ 1 − 1
2
θ2
to be good.
(b) Find cos 10◦
(c) Estimate cos 10◦
57. It follows from part (a) of Example 2 that if θ is small (near
zero) and measured in radians, then one should expect the
approximation
tan θ ≈ θ
to be good.
(a) Find tan 5◦
(b) Find tan 5◦
58. Referring to the accompanying figure, suppose that the an-
gle of elevation of the top of a building, as measured from
a point L feet from its base, is found to be α degrees.
(a) Use the relationship h = L tan α to calculate the height
of a building for which L = 500 ft and α = 6◦
.
(b) Show that if L is large compared to the building height
h, then one should expect good results in approximating
h by h ≈ πLα/180.
(c) Use the result in part (b) to approximate the building
height h in part (a).
ha
L Figure Ex-58
59. (a) Use the Intermediate-Value Theorem to show that the
equation x = cos x has at least one solution in the in-
terval [0, π/2].
(b) Show graphically that there is exactly one solution in
the interval.
(c) Approximate the solution to three decimal places.
60. (a) Use the Intermediate-Value Theorem to show that the
equation x + sin x = 1 has at least one solution in the
interval [0, π/6].
(b) Show graphically that there is exactly one solution in
the interval.
(c) Approximate the solution to three decimal places.
61. In the study of falling objects near the surface of the Earth,
the acceleration g due to gravity is commonly taken to be
9.8 m/s2
or 32 ft/s2
. However, the elliptical shape of the
Earth and other factors cause variations in this constant that
are latitude dependent. The following formula, known as the
Geodetic Reference Formula of 1967, is commonly used to
predict the value of g at a latitude of φ degrees (either north
or south of the equator):
g = 9.7803185(1.0 + 0.005278895 sin2
φ
− 0.000023462 sin4
φ) m/s2
(a) Observe that g is an even function of φ. What does this
suggest about the shape of the Earth, as modeled by the
Geodetic Reference Formula?

Supplementary Exercises 165
(b) Show that g = 9.8 m/s2
somewhere between latitudes
of 38◦
and 39◦
.
62. Let
f(x) =
1 if x is a rational number
0 if x is an irrational number
(a) Make a conjecture about the limit of f(x) as x →0.
(b) Make a conjecture about the limit of xf(x) as x →0.
(c) Prove your conjectures.
SUPPLEMENTARY EXERCISES
the limit if it exists.
(a) lim
x →1
f(x) (b) lim
x →2
f(x) (c) lim
x →3
f(x)
(d) lim
x →4
f(x) (e) lim
x →+ϱ
f(x) (f) lim
x →−ϱ
f(x)
(g) lim
x →3+
f(x) (h) lim
x →3−
f(x) (i) lim
x →0
f(x)
-1 1 2 3 4 5 6 7 8
1
2
3
x
y
Figure Ex-1
2. (a) Find a formula for a rational function that has a verti-
cal asymptote at x = 1 and a horizontal asymptote at
y = 2.
(b) Check your work by using a graphing utility to graph
the function.
3. (a) Write a paragraph or two that describes how the limit of
a function can fail to exist at x = a. Accompany your
description with some specific examples.
(b) Write a paragraph or two that describes how the limit
of a function can fail to exist as x → +ϱ or x → −ϱ.
Also, accompany your description with some specific
examples.
(c) Write a paragraph or two that describes how a function
can fail to be continuous at x = a. Accompany your
description with some specific examples.
4. Show that the conclusion of the Intermediate-Value The-
orem may be false if f is not continuous on the interval
[a, b].
5. In each part, evaluate the function for the stated values of x,
and make a conjecture about the value of the limit. Confirm
your conjecture by finding the limit algebraically.
(a) f(x) =
x − 2
x2 − 4
; lim
x →2+
f(x); x = 2.5, 2.1, 2.01,
2.001, 2.0001, 2.00001
(b) f(x) =
tan 4x
x
; lim
x →0
f(x); x = ±1.0, ±0.1, ±0.01,
±0.001, ±0.0001, ±0.00001
6. In each part, find the horizontal asymptotes, if any.
(a) y =
2x − 7
x2 − 4x
(b) y =
x3
− x2
+ 10
3x2 − 4x
(c) y =
2x2
− 6
x2 + 5x
7. (a) Approximate the value for the limit
lim
x →0
3x
− 2x
x
to three decimal places by constructing an appropriate
table of values.
(b) Confirm your approximation using graphical evidence.
8. According to Ohm’s law, when a voltage of V volts is ap-
plied across a resistor with a resistance of R ohms, a current
of I = V /R amperes flows through the resistor.
(a) How much current flows if a voltage of 3.0 volts is ap-
plied across a resistance of 7.5 ohms?
(b) If the resistance varies by ±0.1 ohm, and the voltage
remains constant at 3.0 volts, what is the resulting range
of values for the current?
(c) If temperature variations cause the resistance to vary
by ±δ from its value of 7.5 ohms, and the voltage re-
mains constant at 3.0 volts, what is the resulting range
of values for the current?
(d) If the current is not allowed to vary by more than
= ±0.001 ampere at a voltage of 3.0 volts, what vari-
ation of ±δ from the value of 7.5 ohms is allowable?
(e) Certain alloys become superconductors as their tem-
perature approaches absolute zero (−273◦
C), meaning
that their resistance approaches zero. If the voltage re-
mains constant, what happens to the current in a super-
conductor as R →0+
?
9. Suppose that f is continuous on the interval [0, 1] and that
0 ≤ f(x) ≤ 1 for all x in this interval.
(a) Sketch the graph of y = x together with a possible
graph for f over the interval [0, 1].
(b) Use the Intermediate-Value Theorem to help prove that
there is at least one number c in the interval [0, 1] such
that f(c) = c.
10. Use algebraic methods to find
(a) lim
θ →0
tan
1 − cos θ
θ
(b) lim
t →1
t − 1
√
t − 1
(c) lim
x →+ϱ
(2x − 1)5
(3x2 + 2x − 7)(x3 − 9x)
(d) lim
θ →0
cos
sin(θ + π)
2θ
.

11. Suppose that f is continuous on the interval [0, 1], that
f(0) = 2, and that f has no zeros in the interval. Prove that
f(x) > 0 for all x in [0, 1].
12. Suppose that
f(x) =
−x4
+ 3, x ≤ 2
x2
+ 9, x > 2
Is f continuous everywhere? Justify your conclusion.
13. Show that the equation x4
+ 5x3
+ 5x − 1 = 0 has at least
two real solutions in the interval [−6, 2].
14. Use the Intermediate-Value Theorem to approximate
√
11
to three decimal places, and check your answer by finding
the root directly with a calculating utility.
15. Suppose that f is continuous at x0 and that f(x0) > 0. Give
either an -δ proof or a convincing verbal argument to show
that there must be an open interval containing x0 on which
f(x) > 0.
16. Sketch the graph of f(x) = |x2
− 4|/(x2
− 4).
17. In each part, approximate the discontinuities of f to three
decimal places.
(a) f(x) =
x + 1
x2 + 2x − 5
(b) f(x) =
x + 3
|2 sin x − x|
18. In Example 3 of Section 2.6 we used the Squeezing Theorem
to prove that
lim
x →0
x sin
1
x
= 0
Why couldn’t we have obtained the same result by writing
lim
x →0
x sin
1
x
= lim
x →0
x · lim
x →0
sin
1
x
= 0 · lim
x →0
sin
1
x
= 0?
In Exercises 19 and 20, find lim
x →a
f(x), if it exists, for
a = 0, 5+
, −5−
, −5, 5, −ϱ, +ϱ
19. (a) f(x) =
√
5 − x (b) f(x) = (x2
−25)/(x−5)
20. (a) f(x) = (x + 5)/(x2
− 25)
(b) f(x) =
(x − 5)/|x − 5|, x = 5
0, x = 5
In Exercises 21–28, find the indicated limit, if it exists.
21. lim
x →0
tan ax
sin bx
(a = 0, b = 0)
22. lim
x →0
sin 3x
tan 3x
23. lim
θ →0
sin 2θ
θ2
24. lim
x →0
x sin x
1 − cos x
25. lim
x →0+
sin x
√
x
26. lim
x →0
sin2
(kx)
x2
, k = 0
27. lim
x →0
3x − sin(kx)
x
, k = 0
28. lim
x →+ϱ
2x + x sin 3x
5x2 − 2x + 1
29. One dictionary describes a continuous function as “one
whose value at each point is closely approached by its values
at neighboring points.”
(a) How would you explain the meaning of the terms
“neighboring points” and “closely approached” to a
nonmathematician?
(b) Write a paragraph that explains why the dictionary def-
inition is consistent with Definition 2.5.1.
30. (a) Show by rationalizing the numerator that
lim
x →0
x2 + 4 − 2
x2
=
1
4
(b) Evaluate f(x) for
x = ±1.0, ±0.1, ±0.01, ±0.001, ±0.0001, ±0.00001
and explain why the values are not getting closer and
closer to the limit.
(c) The accompanying figure shows the graph of f gen-
erated with a graphing utility and zooming in on the
origin. Explain what is happening.
[–0.5, 0.5] × [–0.1, 0.5]
xScl = 0.1, yScl = 0.1
[–5, 5] × [–0.1, 0.5]
xScl = 1, yScl = 0.1
[–5 × 10–6
, 5 × 10–6
] × [–0.1, 0.5]
xScl = 10–6
, yScl = 0.1
Figure Ex-30
In Exercises 31–36, approximate the limit of the function
by looking at its graph and calculating values for some ap-
propriate choices of x. Compare your answer with the value
produced by a CAS.

Supplementary Exercises 167
C 31. lim
x →0
(1 + x)1/x
C 32. lim
x →3
2x
− 8
x − 3
C 33. lim
x →1
sin x − sin 1
x − 1
C 34. lim
x →0+
x−2
(1.001)−1/x
C 35. lim
x →+ϱ
x +
√
x −
√
x
C 36. lim
x →+ϱ
3x
+ 5x 1/x
37. The limit
lim
x →0
sin x
x
= 1
ensures that there is a number δ such that
sin x
x
− 1 < 0.001
if 0 < |x| < δ. Estimate the largest such δ.
38. If $1000 is invested in an account that pays 7% interest
compounded n times each year, then in 10 years there will
be 1000(1 + 0.07/n)10n
dollars in the account. How much
money will be in the account in 10 years if the interest is
compounded quarterly (n = 4)? Monthly (n = 12)? Daily
(n = 365)? Estimate the amount of money that will be in
the account in 10 years if the interest is compounded con-
tinuously, that is, as n→+ϱ?
39. There are various numerical methods other than the method
discussed in Section 2.5 to obtain approximate solutions of
equations of the form f(x) = 0. One such method requires
that the equation be expressed in the form x = g(x), so that
a solution x = c can be interpreted as the value of x where
the line y = x intersects the curve y = g(x), as shown in the
accompanying ﬁgure. If x1 is an initial estimate of c and the
graph of y = g(x) is not too steep in the vicinity of c, then
a better approximation can be obtained from x2 = g(x1)
(see the ﬁgure). An even better approximation is obtained
from x3 = g(x2), and so forth. The formula xn+1 = g(xn)
for n = 1, 2, 3, . . . generates successive approximations
x2, x3, x4, . . . that get closer and closer to c.
(a) The equation x3
−x −1 = 0 has only one real solution.
Show that this equation can be written as
x = g(x) =
3√
x + 1
(b) Graph y = x and y = g(x) in the same coordinate
system for −1 ≤ x ≤ 3.
(c) Starting with an arbitrary estimate x1, make a sketch
that shows the location of the successive iterates
x2 = g(x1), x3 = g(x2), . . .
(d) Use x1 = 1 and calculate x2, x3, . . . , continuing until
you obtain two consecutive values that differ by less
than 10−4
. Experiment with other starting values such
as x1 = 2 or x1 = 1.5.
x
y
y = x y = g(x)
c x3 x2 x1
Figure Ex-39
40. The method described in Exercise 39 will not always work.
(a) The equation x3
− x − 1 = 0 can be expressed as
x = g(x) = x3
− 1. Graph y = x and y = g(x) in
the same coordinate system. Starting with an arbitrary
estimate x1, make a sketch illustrating the locations of
the successive iterates x2 = g(x1), x3 = g(x2), . . . .
(b) Use x1 = 1 and calculate the successive iterates xn for
n = 2, 3, 4, 5, 6.
In Exercises 41 and 42, use the method of Exercise 39 to
approximate the roots of the equation.
41. x5
− x − 2 = 0 42. x − cos x = 0

Ch02

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