lesson 4: continuity

Section 1.5Continuity

V63.0121.002.2010Su, Calculus I

New York University

May 20, 2010

Announcements

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

. . . . . .

Announcements

I Office Hours: MR5:00–5:45, TW 7:50–8:30,CIWW 102 (here)

I Quiz 1 Thursday on1.1–1.4

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

. . . . . .

Objectives

I Understand and apply thedefinition of continuity for afunction at a point or on aninterval.

I Given a piecewise definedfunction, decide where it iscontinuous ordiscontinuous.

I State and understand theIntermediate ValueTheorem.

I Use the IVT to show that afunction takes a certainvalue, or that an equationhas a solution

. . . . . .

Last time

DefinitionWe write

limx→a

f(x) = L

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 aswe like) by taking x to be sufficiently close to a (on either side of a) butnot equal to a.

. . . . . .

Limit Laws for arithmetic

Theorem (Basic Limits)

I limx→a

Theorem (Limit Laws)

Let f and g be functions with limits at a point a. ThenI lim

x→a(f(x) + g(x)) = lim

x→af(x) + lim

x→ag(x)

I limx→a

(f(x)− g(x)) = limx→a

f(x)− limx→a

I limx→a

(f(x) · g(x)) = limx→a

f(x) · limx→a

I limx→a

f(x)g(x)

=limx→a f(x)limx→a g(x)

if limx→a

g(x) ̸= 0

. . . . . .

Hatsumon

Here are some discussion questions to start.

True or FalseAt some point in your life you were exactly three feet tall.

True or FalseAt some point in your life your height (in inches) was equal to yourweight (in pounds).

True or FalseRight now there are a pair of points on opposite sides of the worldmeasuring the exact same temperature.

. . . . . .

Hatsumon

. . . . . .

Hatsumon

. . . . . .

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

limx→a

f(x) = f(a)

This property is so useful it’s worth naming.

. . . . . .

Definition of Continuity

Definition

I Let f be a function definednear a. We say that f iscontinuous at a if

limx→a

f(x) = f(a).

I A function f is continuousif it is continuous at everypoint in its domain. . .x

. . . . . .

Definition of Continuity

Definition

I Let f be a function definednear a. We say that f iscontinuous at a if

limx→a

f(x) = f(a).

I A function f is continuousif it is continuous at everypoint in its domain. . .x

. . . . . .

Scholium

DefinitionLet f be a function defined near a. We say that f is continuous at a if

limx→a

f(x) = f(a).

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

. . . . . .

Free Theorems

Theorem

(a) Any polynomial is continuous everywhere; that is, it is continuouson R = (−∞,∞).

(b) Any rational function is continuous wherever it is defined; that is, itis continuous on its domain.

. . . . . .

Showing a function is continuous.

Example

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

SolutionWe want to show that lim

x→2f(x) = f(2). We have

limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).

Each step comes from the limit laws.

QuestionAt which other points is f continuous?

AnswerThe function f is continuous on (−1/4,∞). It is right continuous at −1/4since lim

x→−1/4+f(x) = f(−1/4).

. . . . . .

Example

limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).

x→−1/4+f(x) = f(−1/4).

. . . . . .

Example

limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).

x→−1/4+f(x) = f(−1/4).

. . . . . .

At which other points?

For reference: f(x) =√4x+ 1

I If a > −1/4, then limx→a

(4x+ 1) = 4a+ 1 > 0, so

limx→a

f(x) = limx→a

√4x+ 1 =

√limx→a

(4x+ 1) =√4a+ 1 = f(a)

and f is continuous at a.

I If a = −1/4, then 4x+ 1 < 0 to the left of a, which means√4x+ 1

is undefined. Still,

limx→a+

f(x) = limx→a+

√4x+ 1 =

√lim

x→a+(4x+ 1) =

√0 = 0 = f(a)

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

. . . . . .

At which other points?

For reference: f(x) =√4x+ 1

I If a > −1/4, then limx→a

(4x+ 1) = 4a+ 1 > 0, so

limx→a

f(x) = limx→a

√4x+ 1 =

√limx→a

(4x+ 1) =√4a+ 1 = f(a)

and f is continuous at a.I If a = −1/4, then 4x+ 1 < 0 to the left of a, which means

√4x+ 1

is undefined. Still,

limx→a+

f(x) = limx→a+

√4x+ 1 =

√lim

x→a+(4x+ 1) =

√0 = 0 = f(a)

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

. . . . . .

Showing a function is continuous

Example

limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).

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

It is right continuous at−1/4 since lim

x→−1/4+f(x) = f(−1/4).

. . . . . .

Showing a function is continuous

Example

limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).

AnswerThe function f is continuous on (−1/4,∞). It is right continuous at−1/4 since lim

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

. . . . . .

The Limit Laws give Continuity Laws

TheoremIf f(x) and g(x) are continuous at a and c is a constant, then thefollowing functions are also continuous at a:

I (f+ g)(x)I (f− g)(x)I (cf)(x)I (fg)(x)

Ifg(x) (if g(a) ̸= 0)

. . . . . .

Why a sum of continuous functions is continuous

We want to show that

limx→a

(f+ g)(x) = (f+ g)(a).

We just follow our nose:

limx→a

(f+ g)(x) = limx→a

[f(x) + g(x)] (def of f+ g)

= limx→a

f(x) + limx→a

g(x) (if these limits exist)

= f(a) + g(a) (they do; f and g are cts.)= (f+ g)(a) (def of f+ g again)

. . . . . .

Trigonometric functions are continuous

I sin and cos are continuous onR.

I tan =sincos

and sec =1cos

arecontinuous on their domain,which isR \

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

arecontinuous on their domain,which is R \ { kπ | k ∈ Z }.

.tan .sec

.cot .csc

. . . . . .

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

.tan .sec

.cot .csc

. . . . . .

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

.cot .csc

. . . . . .

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

.tan .sec

.cot .csc

. . . . . .

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

.tan .sec

. . . . . .

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

.tan .sec

.cot .csc

. . . . . .

Exponential and Logarithmic functions are continuous

For any base a > 1,

I the function x 7→ ax iscontinuous on R

I the function loga iscontinuous on its domain:(0,∞)

I In particular ex andln = loge are continuouson their domains

.loga x

. . . . . .

For any base a > 1,

.loga x

. . . . . .

For any base a > 1,

.loga x

. . . . . .

Inverse trigonometric functions are mostly continuous

I sin−1 and cos−1 are continuous on (−1,1), left continuous at 1,and right continuous at −1.

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

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

.−π

.−π/2

.sin−1

.cos−1.sec−1

..csc−1

.tan−1

.cot−1

. . . . . .

.−π

.−π/2

.sin−1.

.cos−1

.sec−1

..csc−1

.tan−1

.cot−1

. . . . . .

.−π

.−π/2

.sin−1.

.cos−1.sec−1

.csc−1

.tan−1

.cot−1

. . . . . .

.−π

.−π/2

.sin−1.

.cos−1.sec−1

..csc−1

.tan−1

.cot−1

. . . . . .

.−π

.−π/2

.sin−1.

.cos−1.sec−1

..csc−1

.tan−1

.cot−1

. . . . . .

.−π

.−π/2

.sin−1.

.cos−1.sec−1

..csc−1

.tan−1

.cot−1

. . . . . .

What could go wrong?

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

limx→a

f(x) = f(a)

. . . . . .

Continuity FAIL

: The limit does not exist

Example

f(x) =

{x2 if 0 ≤ x ≤ 12x if 1 < x ≤ 2

At which points is f continuous?

SolutionAt any point a in [0,2] besides 1, lim

x→af(x) = f(a) because f is represented by a

polynomial near a, and polynomials have the direct substitution property.However,

limx→1−

f(x) = limx→1−

x2 = 12 = 1

limx→1+

f(x) = limx→1+

2x = 2(1) = 2

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

. . . . . .

Continuity FAIL: The limit does not exist.

Example

f(x) =

{x2 if 0 ≤ x ≤ 12x if 1 < x ≤ 2

SolutionAt any point a in [0,2] besides 1, lim

x→af(x) = f(a) because f is represented by a

polynomial near a, and polynomials have the direct substitution property.However,

limx→1−

f(x) = limx→1−

x2 = 12 = 1

limx→1+

f(x) = limx→1+

2x = 2(1) = 2

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

. . . . . .

Graphical Illustration of Pitfall #1

..−1

. . . . . .

Continuity FAIL

: The function has no value

Example

f(x) =x2 + 2x+ 1

x+ 1At which points is f continuous?

SolutionBecause 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.

. . . . . .

Continuity FAIL: The function has no value

Example

f(x) =x2 + 2x+ 1

x+ 1At which points is f continuous?

SolutionBecause 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.

. . . . . .

...−1

f cannot be continuous where it has no value.

. . . . . .

Continuity FAIL

: function value ̸= limit

Example

f(x) =

{7 if x ̸= 1π if x = 1

Solutionf is not continuous at 1 because f(1) = π but lim

x→1f(x) = 7.

. . . . . .

Continuity FAIL: function value ̸= limit

Example

f(x) =

{7 if x ̸= 1π if x = 1

Solutionf is not continuous at 1 because f(1) = π but lim

x→1f(x) = 7.

. . . . . .

Special types of discontinuites

removable discontinuity The limit limx→a

f(x) exists, but f is not definedat a or its value at a is not equal to the limit at a.

Byre-defining f(a) = lim

x→af(x), f can be made continuous at a

jump discontinuity The limits limx→a−

f(x) and limx→a+

f(x) exist, but aredifferent.

The function cannot be made continuous bychanging a single value.

. . . . . .

Graphical representations of discontinuities

..Presto! continuous!

removable

. .continuous?.

. .continuous?

. . . . . .

removable

. .continuous?.

. .continuous?

. . . . . .

removable

. .continuous?

. . . . . .

removable

. .continuous?

. . . . . .

removable

. .continuous?

. . . . . .

f(x) exists, but f is not definedat a or its value at a is not equal to the limit at a. Byre-defining f(a) = lim

f(x) and limx→a+

f(x) exist, but aredifferent.

The function cannot be made continuous bychanging a single value.

. . . . . .

f(x) exists, but f is not definedat a or its value at a is not equal to the limit at a. Byre-defining f(a) = lim

f(x) and limx→a+

f(x) exist, but aredifferent. The function cannot be made continuous bychanging a single value.

. . . . . .

The greatest integer function

[[x]] is the greatest integer ≤ x.

x [[x]]0 01 1

1.5 11.9 12.1 2

−0.5 −1−0.9 −1−1.1 −2

..−2

..−1

. ..y = [[x]]

This function has a jump discontinuity at each integer.

. . . . . .

The greatest integer function

[[x]] is the greatest integer ≤ x.

x [[x]]0 01 1

1.5 11.9 12.1 2

−0.5 −1−0.9 −1−1.1 −2

..−2

..−1

. ..y = [[x]]

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 beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.

. . . . . .

Illustrating the IVT

. . . . . .

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

and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.

. . . . . .

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

and let N beany number between f(a) and f(b), where f(a) ̸= f(b). Then thereexists a number c in (a,b) such that f(c) = N.

. . . . . .

Suppose that f is continuous on the closed interval [a,b] and let N beany number between f(a) and f(b), where f(a) ̸= f(b).

Then thereexists a number c in (a,b) such that f(c) = N.

. . . . . .

.c3V63.0121.002.2010Su, Calculus I (NYU) Section 1.5 Continuity May 20, 2010 33 / 46

. . . . . .

What the IVT does not say

The Intermediate Value Theorem is an “existence” theorem.I It does not say how many such c exist.I It also does not say how to find c.

Still, it can be used in iteration or in conjunction with other theorems toanswer these questions.

. . . . . .

Using the IVT

Example

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

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

. . . . . .

Using the IVT

Example

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

Note f(1) = 1 andf(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

f(c) = c2 = 2.

. . . . . .

Using the IVT

Example

f(c) = c2 = 2.

. . . . . .

Finding√2 by bisections

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

Using the IVT

Example

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

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

(Morecareful analysis yields 1.32472.)

. . . . . .

Using the IVT

Example

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

Solutionf(1) = −1 and f(2) = 5. So there is a zero between 1 and 2. (Morecareful analysis yields 1.32472.)

. . . . . .

Outline

Continuity

. . . . . .

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

True or FalseAt one point in your life your height in inches equaled your weight inpounds.

True or FalseRight now there are two points on opposite sides of the Earth withexactly the same temperature.

. . . . . .

Question 1: True!

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

. . . . . .

Question 2: True!

I Let h(t) be height in inches and w(t) be weight in pounds, bothvarying continuously over time.

I Let f(t) = h(t)− w(t).I For most of us (call your mom), f(birth) > 0 and f(now) < 0.I So by the IVT there is a point c in (birth,now) where f(c) = 0.I In other words,

h(c)− w(c) = 0 ⇐⇒ h(c) = w(c).

. . . . . .

Question 3

I Let T(θ) be the temperature at the point on the equator atlongitude θ.

I How can you express the statement that the temperature onopposite sides is the same?

I How can you ensure this is true?

. . . . . .

Question 3: True!

I Let f(θ) = T(θ)− T(θ + 180◦)I Then

f(0) = T(0)− T(180)

whilef(180) = T(180)− T(360) = −f(0)

I So somewhere between 0 and 180 there is a point θ wheref(θ) = 0!

. . . . . .

What have we learned today?

I Definition: a function is continuous at a point if the limit of thefunction at that point agrees with the value of the function at thatpoint.

I We often make a fundamental assumption that functions we meetin nature are continuous.

I The Intermediate Value Theorem is a basic property of realnumbers that we need and use a lot.

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