Lesson 5: Continuity (Section 41 slides)

Section 1.5
Continuity

V63.0121.041, Calculus I

New York University

September 20, 2010

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V63.0121.041, Calculus I (NYU) Section 1.5 Continuity September 20, 2010 2 / 47

Grader’s corner

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.


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
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

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) = f (−1/4).
V63.0121.041, 1/4+
x→−Calculus I (NYU) Section 1.5 Continuity September 20, 2010 15 / 47

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) (fg )(x)
(f − g )(x) f
(x) (if g (a) = 0)
(cf )(x) 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.

sin




cos

sin



tan

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



tan sec

sin 1
tan = and sec = are
cos cos cos
which is
π sin
R + kπ k ∈ Z .
2



tan sec

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

cot



tan sec

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

cot csc


Exponential and Logarithmic functions are
continuous

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


continuous

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


continuous

the function x → ax is loga x
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
sin−1 and cos−1 are continuous on (−1, 1), left continuous at 1, and
right continuous at −1.

π

π/2

sin−1

V63.0121.041, Calculus I (NYU) −π/2
Section 1.5 Continuity September 20, 2010 20 / 47

continuous

π

cos−1
π/2

sin−1


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

π

cos−1 sec−1
π/2

sin−1


continuous

π

cos−1 sec−1
π/2

csc−1
sin−1


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

cos−1 sec−1
π/2
tan−1
csc−1
sin−1


continuous
tan−1 and cot−1 are continuous on R.
π
cot−1
cos−1 sec−1
π/2
tan−1
csc−1
sin−1


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 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

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

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


Continuity FAIL: The function has no value

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.



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

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

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

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



y
y

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



y
y

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



y
y

Presto! continuous!
7
2
continuous?
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)

a b x


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

f (x)

f (b)

N

f (a)

a b x


f (x)

f (b)

N

f (a)

a c b x


f (x)

f (b)

N

f (a)

a b x


f (x)

f (b)

N

f (a)

a c1 c2 c3 b x


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 to ﬁnd zeroes

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


Using the IVT to ﬁnd zeroes

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.


Finding a zero by bisection

y

x f (x)

x



y

x f (x)
1 −1

x



y

x f (x)
1 −1

2 5

x



y

x f (x)
1 −1

1.5 0.875
2 5

x



y

x f (x)
1 −1
1.25 − 0.296875

1.5 0.875
2 5

x



y

x f (x)
1 −1
1.25 − 0.296875

1.375 0.224609
1.5 0.875
2 5

x



y

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

x



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

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



Example

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



Example

Proof.
Let f (x) = x 2 , 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) = c 2 = 2.


Outline

Continuity





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


Question 1

Answer
The answer is 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

Answer
The answer is 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

Answer
The answer is TRUE.
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

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 5: Continuity (Section 41 slides)

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