Lesson 5: Continuity (Section 21 handout)

V63.0121.021, Calculus I Section 1.5 : Continuity September 20, 2010

Notes
Section 1.5
Continuity

V63.0121.021, Calculus I

New York University

September 20, 2010

Announcements
Notes

Oﬃce Hours: Tuesday,
Wednesday, 3pm–4pm
TAs have oﬃce hours on
website

V63.0121.021, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 2 / 47

Grader’s corner
Notes

HW Grades will be on
blackboard this week, and
the papers will be returned
in recitation
Remember units when
computing slopes
Remember to staple your
papers—you have been
warned.


1


Objectives
Notes

Understand and apply the
definition of continuity for a
function at a point or on an
interval.
Given a piecewise defined
function, decide where it is
continuous or discontinuous.
State and understand the
Intermediate Value
Theorem.
Use the IVT to show that a
function takes a certain
value, or that an equation
has a solution


Last time
Notes

Definition
We write
lim f (x) = L
x→a

and say

“the limit of f (x), as x approaches a, equals L”

if we can make the values of f (x) arbitrarily close to L (as close to L as we
like) by taking x to be sufficiently close to a (on either side of a) but not
equal to a.


Limit Laws for arithmetic
Notes
Theorem (Basic Limits)

lim x = a
x→a
lim c = c
x→a

Theorem (Limit Laws)
Let f and g be functions with limits at a point a. Then
lim (f (x) + g (x)) = lim f (x) + lim g (x)
x→a x→a x→a
lim (f (x) − g (x)) = lim f (x) − lim g (x)
x→a x→a x→a
lim (f (x) · g (x)) = lim f (x) · lim g (x)
x→a x→a x→a
f (x) limx→a f (x)
lim = if lim g (x) = 0
x→a g (x) limx→a g (x) x→a


2


Hatsumon
Notes

Here are some discussion questions to start.
True or False
At some point in your life you were exactly three feet tall.

True or False
At some point in your life your height (in inches) was equal to your weight
(in pounds).

True or False
Right now there are a pair of points on opposite sides of the world
measuring the exact same temperature.


Outline
Notes

Continuity

The Intermediate Value Theorem

Back to the Questions


Recall: Direct Substitution Property
Notes

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

This property is so useful it’s worth naming.


3


Definition of Continuity
Notes
y
Definition

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

lim f (x) = f (a).
x→a

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


Scholium
Notes

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

There are three important parts to this definition.
The function has to have a limit at a,
the function has to have a value at a,
and these values have to agree.


Free Theorems
Notes

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.


4


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

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.

Question
At which other points is f continuous?

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


At which other points?
Notes
√
For reference: f (x) = 4x + 1
If a > −1/4, then lim (4x + 1) = 4a + 1 > 0, so
x→a
√ √
lim f (x) = lim 4x + 1 = lim (4x + 1) = 4a + 1 = f (a)
x→a x→a x→a

and f is continuous at a.
√
If a = −1/4, then 4x + 1 < 0 to the left of a, which means 4x + 1 is
undeﬁned. Still,
√ √
lim+ f (x) = lim+ 4x + 1 = lim+ (4x + 1) = 0 = 0 = f (a)
x→a x→a x→a

so f is continuous on the right at a = −1/4.


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

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.

Question
At which other points is f continuous?

Answer
The function f is continuous on (−1/4, ∞). It is right continuous at −1/4
since lim f (x) = f (−1/4).
V63.0121.021, 1/4+
x→−Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47

5


The Limit Laws give Continuity Laws
Notes

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


Why a sum of continuous functions is continuous
Notes

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)] (def of f + g )
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) (def of f + g again)


Trigonometric functions are continuous
Notes
tan sec

sin and cos are continuous on R.
sin 1
tan = and sec = are
cos cos cos
continuous on their domain,
which is
π sin
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
continuous on their domain,
which is R { kπ | k ∈ Z }.

cot csc


6


Exponential and Logarithmic functions are
continuous Notes

For any base a > 1, ax
x loga x
the function x → a is
continuous on R
the function loga is
continuous on its domain:
(0, ∞)
In particular e x and
ln = loge are continuous on
their domains


Inverse trigonometric functions are mostly
continuous Notes

sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.
sec−1 and csc−1 are continuous on (−∞, −1) ∪ (1, ∞), left
continuous at −1, and right continuous at 1.
tan−1 and cot−1 are continuous on R.
π
cot−1
cos−1 sec−1
π/2
tan−1
csc−1
−1
sin

V63.0121.021, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47

−π

What could go wrong?
Notes

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


7


Continuity FAIL: The limit does not exist
Notes
Example
Let
x2 if 0 ≤ x ≤ 1
f (x) =
2x if 1 < x ≤ 2
At which points is f continuous?

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

lim f (x) = lim x 2 = 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
Notes

y
4

3
The function cannot be
2 continuous at a point if the
function has no limit at that
point.
1

x
−1 1 2
−1


Continuity FAIL: The function has no value
Notes

Example
Let
x 2 + 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.


8


Notes

y

1 The function cannot be
continuous at a point outside its
domain (that is, a point where it
x has no value).
−1


Continuity FAIL: function value = limit
Notes

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


Notes

y

7 If the function has a limit and a
value at a point the two must
π still agree.

x
1


9


Special types of discontinuites
Notes

removable discontinuity The limit lim f (x) exists, but f is not defined
x→a
at a or its value at a is not equal to the limit at a. By
re-defining f (a) = lim f (x), f can be made continuous at a
x→a
jump discontinuity The limits lim f (x) and lim+ f (x) exist, but are
x→a− x→a
different. The function cannot be made continuous by
changing a single value.


Graphical representations of discontinuities
Notes

y
y

Presto! continuous!
7
2 continuous?
continuous?
π
1 continuous?
x
x
1
1
removable
jump


Special types of discontinuites
Notes

removable discontinuity The limit lim f (x) exists, but f is not defined
x→a
at a or its value at a is not equal to the limit at a. By
re-defining f (a) = lim f (x), f can be made continuous at a
x→a
jump discontinuity The limits lim f (x) and lim+ f (x) exist, but are
x→a− x→a
different. The function cannot be made continuous by
changing a single value.


10


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

3
x [[x]] y = [[x]]
0 0 2
1 1
1.5 1 1
1.9 1
2.1 2 x
−0.5 −1 −2 −1 1 2 3
−0.9 −1 −1
−1.1 −2
−2
This function has a jump discontinuity at each integer.


Outline
Notes

Continuity




A Big Time Theorem
Notes

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.


11


Illustrating the IVT
Notes
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.
f (x)

f (b)

N

f (a)

a c1 c c2 c3 b x


What the IVT does not say
Notes

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 to ﬁnd zeroes
Notes

Example
Let f (x) = x 3 − x − 1. Show that there is a zero for f on the interval [1, 2].

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

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


12


Finding a zero by bisection
Notes

y

x f (x)
1 −1
1.25 − 0.296875
1.3125 − 0.0515137
1.375 0.224609
1.5 0.875
2 5

(More careful analysis yields x
1.32472.)


Using the IVT to assert existence of numbers
Notes

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


Outline
Notes

Continuity




13


Notes

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

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 1
Notes

To be discussed in class!


Question 2
Notes


14


Question 3
Notes


Question 3
Notes


What have we learned today?
Notes

Deﬁnition: a function is continuous at a point if the limit of the
function at that point agrees with the value of the function at that
point.
We often make a fundamental assumption that functions we meet in
nature are continuous.
The Intermediate Value Theorem is a basic property of real numbers
that we need and use a lot.


15

Lesson 5: Continuity (Section 21 handout)

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