Lesson 5: Continuity

Section 1.5
Continuity

V63.0121.034, Calculus I

September 16, 2009

Announcements
Please put your homework in the envelope by last name
Quiz next week in Recitation on §§1.1–1.3
WebAssignments 1 and 2 due Monday
. . . . . .

Hatsumon

Here are some discussion questions to start.
Were you ever exactly three feet tall?

. . . . . .

Hatsumon

Was your height (in inches) ever equal to your weight (in
pounds)?

. . . . . .

Hatsumon

Was your height (in inches) ever equal to your weight (in
pounds)?
Is there a pair of points on opposite sides of the world at the
same temperature at the same time?

. . . . . .

Outline

Continuity

The Intermediate Value Theorem

Back to the Questions

. . . . . .

Recall: Direct Substitution Property

Theorem (The Direct Substitution Property)
If f is a polynomial or a rational function and a is in the domain of
f, then
lim f(x) = f(a)
x→a

. . . . . .

Definition of Continuity
Definition
Let f be a function
defined near a. We say
that f is continuous at a
if

lim f(x) = f(a).
x→a

. . . . . .

Definition of Continuity
Definition y
.
Let f be a function
defined near a. We say
that f is continuous at a f
. (a ) .
if

lim f(x) = f(a).
x→a

A function f is
continuous if it is
continuous at every . x
.
point in its domain. a
.

There are three important parts to this definition.
The limit has to exist
the function has to be defined
and these values have to agree. . . . . . .

Free Theorems

Theorem
(a) Any polynomial is continuous everywhere; that is, it is
continuous on R = (−∞, ∞).
(b) Any rational function is continuous wherever it is deﬁned;
that is, it is continuous on its domain.

. . . . . .

Showing a function is continuous
Example
√
Let f(x) = 4x + 1. Show that f is continuous at 2.

. . . . . .

Example
√

Solution
We want to show that lim f(x) = f(2). We have
x→2
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2).
x→a x→2 x→2

Each step comes from the limit laws.

. . . . . .

Example
√

Solution
x→2
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2).
x→a x→2 x→2


Question
At which other points is f continuous?

. . . . . .

Example
√

Solution
x→2
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2).
x→a x→2 x→2


Question

Answer
The function f is continuous on (−1/4, ∞).

. . . . . .

Example
√

Solution
x→2
√ √ √
lim f(x) = lim 4x + 1 = lim (4x + 1) = 9 = 3 = f(2).
x→a x→2 x→2


Question

Answer
The function f is continuous on (−1/4, ∞). It is right continuous
at the point −1/4 since lim f(x) = f(−1/4).
x→−1/4+
. . . . . .

The Limit Laws give Continuity Laws

Theorem
If f and g are continuous at a and c is a constant, then the
following functions are also continuous at a:
f+g
f−g
cf
fg
f
(if g(a) ̸= 0)
g

. . . . . .

Why a sum of continuous functions is continuous

We want to show that

lim (f + g)(x) = (f + g)(a).
x→a

We just follow our nose:

lim (f + g)(x) = lim [f(x) + g(x)]
x→a x→a
= lim f(x) + lim g(x) (if these limits exist)
x→a x→a
= f(a) + g(a) (they do; f and g are cts)
= (f + g)(a)

. . . . . .

Trigonometric functions are continuous

sin and cos are continuous
on R.

.
s
. in

. . . . . .


on R.

c
. os

.
s
. in

. . . . . .


t
.an
on R.
sin 1
tan = and sec =
cos cos c
. os
are continuous on their
domain, which is
{π } .
R + kπ k ∈ Z . s
. in
2

. . . . . .


t
.an s
. ec
on R.
sin 1
tan = and sec =
cos cos c
. os
domain, which is
{π } .
R + kπ k ∈ Z . s
. in
2

. . . . . .


t
.an s
. ec
on R.
sin 1
tan = and sec =
cos cos c
. os
domain, which is
{π } .
R + kπ k ∈ Z . s
. in
2
cos 1
cot = and csc =
sin sin
domain, which is
R { kπ | k ∈ Z }.
c
. ot

. . . . . .


t
.an s
. ec
on R.
sin 1
tan = and sec =
cos cos c
. os
domain, which is
{π } .
R + kπ k ∈ Z . s
. in
2
cos 1
cot = and csc =
sin sin
domain, which is
R { kπ | k ∈ Z }.
c
. ot . sc
c

. . . . . .

Exponential and Logarithmic functions are continuous

For any base a > 1, .x
a
the function x → ax is
continuous on R

.

. . . . . .


a
the function x → ax is .oga x
l
continuous on R
the function loga is
continuous on its .
domain: (0, ∞)

. . . . . .


a
the function x → ax is .oga x
l
continuous on R
the function loga is
continuous on its .
domain: (0, ∞)
In particular ex and
ln = loge are continuous
on their domains

. . . . . .

Inverse trigonometric functions are mostly continuous
sin−1 and cos−1 are continuous on (−1, 1), left continuous
at 1, and right continuous at −1.

.
π

.
. /2
π

.
. in−1
s

.
−
. π/2
. . . . . .


.
.
π

. os−1
c .
. /2
π

. .
. in−1
s

.
−
. π/2
. . . . . .

sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.

.
.
π

. os−1
c . . ec−1
s
. /2
π

. .
. in−1
s

.
−
. π/2
. . . . . .


.
.
π

. os−1
c . . ec−1
s
. /2
π

. sc−1
c
. .
. in−1
s

.
−
. π/2
. . . . . .

tan−1 and cot−1 are continuous on R.
.
.
π

. os−1
c . . ec−1
s
. /2
π
.an−1
t
. sc−1
c
. .
. in−1
s

.
−
. π/2
. . . . . .

tan−1 and cot−1 are continuous on R.
.
.
π
. ot−1
c
. os−1
c . . ec−1
s
. /2
π
.an−1
t
. sc−1
c
. .
. in−1
s

.
−
. π/2
. . . . . .

What could go wrong?

In what ways could a function f fail to be continuous at a point a?
Look again at the deﬁnition:

lim f(x) = f(a)
x→a

. . . . . .

Pitfall #1
Example
Let {
x2 if 0 ≤ x ≤ 1
f (x ) =
2x if 1 < x ≤ 2
At which points is f continuous?

. . . . . .

Pitfall #1: The limit does not exist
Example
Let {
x2 if 0 ≤ x ≤ 1
f (x ) =
2x if 1 < x ≤ 2

Solution
At any point a in [0, 2] besides 1, lim f(x) = f(a) because f is
x→a
represented by a polynomial near a, and polynomials have the
direct substitution property. However,

lim f(x) = lim x2 = 12 = 1
x→1− x→1−
lim f(x) = lim 2x = 2(1) = 2
x→1+ x→1+

So f has no limit at 1. Therefore f is not continuous at 1.
. . . . . .

Graphical Illustration of Pitfall #1

y
.
. .
4 .

. .
3

. .
2 .

. .
1 .

. . . . x
.
−
. 1 1
. 2
.
. 1 .
−

. . . . . .

Pitfall #2

Example
Let
x2 + 2x + 1
f (x ) =
x+1

. . . . . .

Pitfall #2: The function has no value

Example
Let
x2 + 2x + 1
f (x ) =
x+1

Solution
Because f is rational, it is continuous on its whole domain. Note
that −1 is not in the domain of f, so f is not continuous there.

. . . . . .


y
.

. .
1

. . x
.
−
. 1

f cannot be continuous where it has no value.

. . . . . .

Pitfall #3

Example
Let {
7 if x ̸= 1
f(x) =
π if x = 1

. . . . . .

Pitfall #3: function value ̸= limit

Example
Let {
7 if x ̸= 1
f(x) =
π if x = 1

Solution
f is not continuous at 1 because f(1) = π but lim f(x) = 7.
x→1

. . . . . .


y
.

. .
7 .

. .
π .

. . x
.
1
.

. . . . . .

Special types of discontinuites

removable discontinuity The limit lim f(x) exists, but f is not
x→a
deﬁned at a or its value at a is not equal to the limit
at a.
jump discontinuity The limits lim f(x) and lim f(x) exist, but
x→a− x→a+
are different. f(a) is one of these limits.

. . . . . .

Graphical representations of discontinuities

y
.
. .
4 .
y
.

. .
3
. .
7 .
. .
2 .
. .
π .
. .
1 .
. . x
.
1
. . . . . x
.
−
. 1 1
. 2
.
removable . 1 .
−
jump

. . . . . .

The greatest integer function
[[x]] is the greatest integer ≤ x.
y
.

. .
3
y
. = [[x]]
. .
2 . .

. .
1 . .

. . . . . . x
.
−
. 2 −
. 1 1
. 2
. 3
.
.. 1 .
−

. .. 2 .
−

. . . . . .

The greatest integer function
[[x]] is the greatest integer ≤ x.
y
.

. .
3
y
. = [[x]]
. .
2 . .

. .
1 . .

. . . . . . x
.
−
. 2 −
. 1 1
. 2
. 3
.
.. 1 .
−

. .. 2 .
−

This function has a jump discontinuity at each integer.
. . . . . .

A Big Time Theorem

Theorem (The Intermediate Value Theorem)
Suppose that f is continuous on the closed interval [a, b] and let N
be any number between f(a) and f(b), where f(a) ̸= f(b). Then
there exists a number c in (a, b) such that f(c) = N.

. . . . . .

Illustrating the IVT

f
. (x )

. x
.

. . . . . .

Suppose that f is continuous on the closed interval [a, b]

f
. (x )

.

.

. x
.

. . . . . .

Suppose that f is continuous on the closed interval [a, b]

f
. (x )

f
. (b ) .

f
. (a ) .

. a
. x
.
b
.

. . . . . .

be any number between f(a) and f(b), where f(a) ̸= f(b).

f
. (x )

f
. (b ) .

N
.

f
. (a ) .

. a
. x
.
b
.

. . . . . .

f
. (x )

f
. (b ) .

N
. .

f
. (a ) .

. a
. c
. x
.
b
.

. . . . . .

f
. (x )

f
. (b ) .

N
.

f
. (a ) .

. a
. x
.
b
.

. . . . . .

f
. (x )

f
. (b ) .

N
. . . .

f
. (a ) .

. a c
. .1 x
.
c
.2 c b
.3 .

. . . . . .

What the IVT does not say

The Intermediate Value Theorem is an “existence” theorem.
It does not say how many such c exist.
It also does not say how to ﬁnd c.
Still, it can be used in iteration or in conjunction with other
theorems to answer these questions.

. . . . . .

Using the IVT

Example
Suppose we are unaware of the square root function and that it’s
continuous. Prove that the square root of two exists.

. . . . . .

Using the IVT

Example

Proof.
Let f(x) = x2 , a continuous function on [1, 2].

. . . . . .

Using the IVT

Example

Proof.
Let f(x) = x2 , a continuous function on [1, 2]. Note f(1) = 1 and
f(2) = 4. Since 2 is between 1 and 4, there exists a point c in
(1, 2) such that
f(c) = c2 = 2.

. . . . . .

Using the IVT

Example

Proof.
Let f(x) = x2 , a continuous function on [1, 2]. Note f(1) = 1 and
f(2) = 4. Since 2 is between 1 and 4, there exists a point c in
(1, 2) such that
f(c) = c2 = 2.

In fact, we can “narrow in” on the square root of 2 by the method
of bisections.

. . . . . .

√
Finding 2 by bisections

. .(x) = x2
x f

. ..
2 4

. ..
1 1

. . . . . .

√

. .(x) = x2
x f

. ..
2 4

. .5 . . .25
1 2

. ..
1 1

. . . . . .

√

. .(x) = x2
x f

. ..
2 4

. .5 . . .25
1 2

. .25 . . .5625
1 1

. ..
1 1

. . . . . .

√

. .(x) = x2
x f

. ..
2 4

1
. .5 . . .25
2
1
. .375 . . .890625
1
1
. .25 . . .5625
1

. ..
1 1

. . . . . .

√

. .(x) = x2
x f

. ..
2 4

. . .25
1
. .5 . 2.06640625
1
. .4375
1
. .375 . . .890625
2
1
.
1
. .25 . . .5625
1

. ..
1 1

. . . . . .

Using the IVT

Example
Let f(x) = x3 − x − 1. Show that there is a zero for f.

Solution
f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.

. . . . . .

Using the IVT

Example
Let f(x) = x3 − x − 1. Show that there is a zero for f.

Solution
f(1) = −1 and f(2) = 5. So there is a zero between 1 and 2.
(More careful analysis yields 1.32472.)

. . . . . .


True or False
At one point in your life you were exactly three feet tall.

. . . . . .

Question 1: True!

Let h(t) be height, which varies continuously over time. Then
h(birth) < 3 ft and h(now) > 3 ft. So there is a point c in
(birth, now) where h(c) = 3.

. . . . . .


True or False

True or False
At one point in your life your height in inches equaled your
weight in pounds.

. . . . . .

Question 2: True!

Let h(t) be height in inches and w(t) be weight in pounds, both
varying continuously over time. Let f(t) = h(t) − w(t). For most
of us (call your mom), f(birth) > 0 and f(now) < 0. So there is a
point c in (birth, now) where f(c) = 0. In other words,

h(c) − w(c) = 0 ⇐⇒ h(c) = w(c).

. . . . . .


True or False

True or False
At one point in your life your height in inches equaled your
weight in pounds.

True or False
Right now there are two points on opposite sides of the Earth
with exactly the same temperature.

. . . . . .

Question 3

Let T(θ) be the temperature at the point on the equator at
longitude θ.
How can you express the statement that the temperature on
opposite sides is the same?
How can you ensure this is true?

. . . . . .

Question 3: True!

Let f(θ) = T(θ) − T(θ + 180◦ )
Then
f(0) = T(0) − T(180)
while
f(180) = T(180) − T(360) = −f(0)
So somewhere between 0 and 180 there is a point θ where
f(θ) = 0!

. . . . . .

Lesson 5: Continuity

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Lesson 5: Continuity