Lesson 4: Continuity

Section 1.5
Continuity

V63.0121.002.2010Su, Calculus I

New York University

May 20, 2010

Announcements

Office Hours: MR 5:00–5:45, TW 7:50–8:30, CIWW 102 (here)
Quiz 1 Thursday on 1.1–1.4

. . . . . .

Announcements

Office Hours: MR
5:00–5:45, TW 7:50–8:30,
CIWW 102 (here)
Quiz 1 Thursday on
1.1–1.4

. . . . . .

V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 2 / 46

Objectives

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

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

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
. . . . . .


Hatsumon

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

. . . . . .


Hatsumon

True or False

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

. . . . . .


Hatsumon

True or False

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

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

This property is so useful it’s worth naming.

. . . . . .


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

y
.
Definition

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

. . . . . .


Scholium

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

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?

. . . . . . .


At which other points?

√
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.

. . . . . .


At which other points?

√
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.

. . . . . .


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
. . . . . .

−1/4 since lim f(x)
V63.0121.002.2010Su, Calculus I (NYU) = f(−1 Section 1.5 Continuity
/4). May 20, 2010 14 / 46

The Limit Laws give Continuity Laws

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)
(f − g)(x)
(cf)(x)
(fg)(x)
f
(x) (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)] (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

sin and cos are continuous on
R.

.
s
. in

. . . . . .



R.

c
. os

.
s
. in

. . . . . .



t
.an

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

. . . . . .



t
.an s
. ec

R.
sin 1
. os
cos cos
{π
which is } .
s
. in
R + kπ k ∈ Z .
2

. . . . . .



t
.an s
. ec

R.
sin 1
. os
cos cos
{π
which is } .
s
. in
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
which is R { kπ | k ∈ Z }.
c
. ot

. . . . . .



t
.an s
. ec

R.
sin 1
. os
cos cos
{π
which is } .
s
. in
R + kπ k ∈ Z .
2
cos 1
cot = and csc = are
sin sin
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 l
.oga x
continuous on R
the function loga is
continuous on its domain: .
(0, ∞)

. . . . . .



a
the function x → ax is l
.oga x
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

. . . . . .


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

.
. . . . . .


Continuity FAIL: 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 represented by a
x→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 .
−

. . . . . .


Continuity FAIL

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

. . . . . .


Continuity FAIL: 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.
. . . . . .


Continuity FAIL

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

. . . . . .


Continuity FAIL: 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 defined
x→a
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 are
x→a− x→a
different.

. . . . . .


Graphical representations of discontinuities

.
y
.
y
.

. .
7 .
. .
2 .

. .
π .
. .
1 .

. . x
. . . x
.
1
.
1
.
removable
jump

. . . . . .



.
y
.
y
.

P
. resto! continuous!
. .
7 .
. .
2 .

. .
π
. .
1 .

. . x
. . . x
.
1
.
1
.
removable
jump

. . . . . .



.
y
.
y
.

P
. .
7 .
. .
2 .

. .
π
. .
1 . . ontinuous?
c
. . x
. . . x
.
1
.
1
.
removable
jump

. . . . . .



.
y
.
y
.

P
. .
7 .
. .
2 . . ontinuous?
c
. .
π
. .
1 .

. . x
. . . x
.
1
.
1
.
removable
jump

. . . . . .



.
y
.
y
.

P
. .
7 .
. .
2 .
. . ontinuous?
c
. .
π
. .
1 .

. . x
. . . x
.
1
.
1
.
removable
jump

. . . . . .



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
x→a− x→a
different.

. . . . . .



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
x→a− x→a
different. The function cannot be made continuous by
changing a single value.

. . . . . .


The greatest integer function

[[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 .
−

. . . . . .


The greatest integer function

[[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

Continuity



. . . . . .


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) .

. x
.
a
. b
.
. . . . . .


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

f
.(x)

f
.(b) .

N
.

f
.(a) .

. x
.
a
. b
.
. . . . . .


f
.(x)

f
.(b) .

N
. .

f
.(a) .

. x
.
a
. c
. b
.
. . . . . .


f
.(x)

f
.(b) .

N
.

f
.(a) .

. x
.
a
. b
.
. . . . . .


f
.(x)

f
.(b) .

N
. . . .

f
.(a) .

. x
.
ac
. .1 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

1
. .5 . . .25
1
. .4375 . 2.06640625
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.)

. . . . . .


Outline

Continuity



. . . . . .



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 by the IVT 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 by the IVT 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!

. . . . . .


What have we learned today?

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.

. . . . . .


Lesson 4: Continuity

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