lesson 5: continuity (slides)

Sec on 1.5Con nuity

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

February 7, 2011

Announcements

I Get-to-know-you extracredit due FridayFebruary 11

I Quiz 1 February 17/18 inrecita on

ObjectivesI Understand and apply the defini on of

con nuity for a func on at a point oron an interval.

I Given a piecewise defined func on,decide where it is con nuous ordiscon nuous.

I State and understand the IntermediateValue Theorem.

I Use the IVT to show that a func ontakes a certain value, or that anequa on has a solu on

Last timeDefini onWe 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 Las we like) by taking x to be sufficiently close to a (on either side ofa) but not equal to a.

Basic Limits

Theorem (Basic Limits)

I limx→a

x = aI lim

x→ac = c

Limit Laws for arithmeticTheorem (Limit Laws)

Let f and g be func ons 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

HatsumonHere are some discussion ques ons 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.

Outline

Con nuity

The Intermediate Value Theorem

Back to the Ques ons

Recall: Direct SubstitutionProperty

Theorem (The Direct Subs tu on Property)

If f is a polynomial or a ra onal func on 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

Defini on

I Let f be a func on defined neara. We say that f is con nuous ata if

limx→a

f(x) = f(a).

I A func on f is con nuous if it iscon nuous at every point in itsdomain.

Definition of Continuity

Defini on

I Let f be a func on defined neara. We say that f is con nuous ata if

limx→a

f(x) = f(a).

I A func on f is con nuous if it iscon nuous at every point in itsdomain. .. x.

ScholiumDefini onLet f be a func on defined near a. We say that f is con nuous at a if

limx→a

f(x) = f(a).

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

Free Theorems

Theorem

(a) Any polynomial is con nuous everywhere; that is, it iscon nuous on R = (−∞,∞).

(b) Any ra onal func on is con nuous wherever it is defined; that is,it is con nuous on its domain.

Showing a function is continuous..Example

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

Solu onWe 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.

Showing a function is continuous..Example

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

Solu onWe 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.

At which other points?Ques on

As before, let f(x) =√4x+ 1. At which points is f con nuous?

Solu on

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 con nuous at a.

Solu on

I If a = −1/4, then 4x+ 1 < 0 to the le of a, which means√4x+ 1 is undefined. S ll,

limx→a+

f(x) = limx→a+

√4x+ 1 =

√limx→a+

(4x+ 1) =√0 = 0 = f(a)

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

Limit Laws give Continuity Laws

TheoremIf f(x) and g(x) are con nuous at a and c is a constant, then thefollowing func ons are also con nuous at a:

I (f+ g)(x)I (f− g)(x)I (cf)(x)

I (fg)(x)

Ifg(x) (if g(a) ̸= 0)

Sum of continuous is continuousWe want to show that

limx→a

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

We just follow our nose.

(def of f+ g) limx→a

(f+ g)(x) = limx→a

[f(x) + g(x)]

(if these limits exist) = limx→a

f(x) + limx→a

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

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

limx→a

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

(f+ g)(x)

= limx→a

[f(x) + g(x)]

f(x) + limx→a

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

limx→a

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

[f(x) + g(x)]

(if these limits exist)

= limx→a

f(x) + limx→a

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

(they do; f and g are cts.)

= f(a) + g(a)(def of f+ g again) = (f+ g)(a)

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

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

(def of f+ g again) = (f+ g)(a)

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

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

= (f+ g)(a)

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

limx→a

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

[f(x) + g(x)]

f(x) + limx→a

Trig functions are continuousI sin and cos are con nuouson R.

I tan =sincos

and sec =1cos

arecon nuous on their domain,which isR \

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

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

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

I tan =sincos

and sec =1cos

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

Exp and Log are continuousFor any base a > 1,I the func on x 7→ ax iscon nuous on R

I the func on loga iscon nuous on itsdomain: (0,∞)

I In par cular ex andln = loge are con nuouson their domains

loga x

Inverse trigonometric functionsare mostly continuous

I sin−1 and cos−1 are con nuous on (−1, 1), le con nuous at 1,and right con nuous at−1.

I sec−1 and csc−1 are con nuous on (−∞,−1) ∪ (1,∞), lecon nuous at−1, and right con nuous at 1.

I tan−1 and cot−1 are con nuous on R.

−π/2

...sin−1

cos−1

sec−1

... csc−1

... tan−1

cot−1

−π/2

...sin−1 ...

cos−1

sec−1

... csc−1

... tan−1

cot−1

−π/2

...sin−1 ...

cos−1

sec−1

. csc−1

... tan−1

cot−1

−π/2

...sin−1 ...

cos−1

sec−1

... csc−1

. tan−1

cot−1

−π/2

...sin−1 ...

cos−1

sec−1

... csc−1

... tan−1

cot−1

−π/2

...sin−1 ...

cos−1

sec−1

... csc−1

... tan−1

cot−1

What could go wrong?

In what ways could a func on f fail to be con nuous at a point a?Look again at the equa on from the defini on:

limx→a

f(x) = f(a)

Continuity FAIL

: no limit

..Example

Let f(x) =

{x2 if 0 ≤ x ≤ 12x if 1 < x ≤ 2

. At which points is f con nuous?

Solu onAt any point a besides 1, lim

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

polynomial near a, and polynomials have the direct subs tu on property.

limx→1−

f(x) = limx→1−

x2 = 12 = 1 and limx→1+

f(x) = limx→1+

2x = 2(1) = 2

So f has no limit at 1. Therefore f is not con nuous at 1.

Continuity FAIL: no limit..Example

Let f(x) =

{x2 if 0 ≤ x ≤ 12x if 1 < x ≤ 2

. At which points is f con nuous?

Solu onAt any point a besides 1, lim

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

polynomial near a, and polynomials have the direct subs tu on property.

limx→1−

f(x) = limx→1−

x2 = 12 = 1 and limx→1+

f(x) = limx→1+

2x = 2(1) = 2

So f has no limit at 1. Therefore f is not con nuous at 1.

Graphical Illustration of Pitfall #1

..−1

The func on cannot becon nuous at a point if thefunc on has no limit at thatpoint.

..−1

FAILThe func on cannot becon nuous at a point if thefunc on has no limit at thatpoint.

Continuity FAIL

: no value

Example

Letf(x) =

x2 + 2x+ 1x+ 1

At which points is f con nuous?

Solu onBecause f is ra onal, it is con nuous on its whole domain. Note that−1 is not in the domain of f, so f is not con nuous there.

Continuity FAIL: no valueExample

Letf(x) =

x2 + 2x+ 1x+ 1

Solu onBecause f is ra onal, it is con nuous on its whole domain. Note that−1 is not in the domain of f, so f is not con nuous there.

...−1

1 The func on cannot becon nuous at a point outsideits domain (that is, a pointwhere it has no value).

...−1

FAIL The func on cannot becon nuous at a point outsideits domain (that is, a pointwhere it has no value).

Continuity FAIL

: value ̸= limit

Example

f(x) =

{7 if x ̸= 1π if x = 1

Solu onf is not con nuous at 1 because f(1) = π but lim

x→1f(x) = 7.

Continuity FAIL: value ̸= limitExample

f(x) =

{7 if x ̸= 1π if x = 1

Solu onf is not con nuous at 1 because f(1) = π but lim

x→1f(x) = 7.

If the func on has a limit anda value at a point the twomust s ll agree.

...FAIL

If the func on has a limit anda value at a point the twomust s ll agree.

Special types of discontinuities

removable discon nuity The limit limx→a

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

By re-defining f(a) = limx→a

f(x), f can be made con nuousat a

jump discon nuity The limits limx→a−

f(x) and limx→a+

f(x) exist, but aredifferent.

The func on cannot be made con nuous bychanging a single value.

Special discontinuities graphically

Presto! con nuous!

removable

con nuous?

Presto! con nuous!

removable

con nuous?

Presto! con nuous!

removable

......

con nuous?

Presto! con nuous!

removable

con nuous?

Presto! con nuous!

removable

con nuous?

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

x→af(x), f can be made con nuous

at ajump discon nuity The limits lim

x→a−f(x) and lim

x→a+f(x) exist, but are

different.

The func on cannot be made con nuous bychanging a single value.

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

x→af(x), f can be made con nuous

at ajump discon nuity The limits lim

x→a−f(x) and lim

x→a+f(x) exist, but are

different. The func on cannot be made con nuous 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 func on has a jump discon nuity 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 func on has a jump discon nuity at each integer.

Outline

Con nuity

A Big Time Theorem

Theorem (The Intermediate Value Theorem)

Suppose that f is con nuous 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

Theorem

Suppose that f is con nuouson the closed interval [a, b]and let N be any numberbetween f(a) and f(b), wheref(a) ̸= f(b). Then there existsa number c in (a, b) such thatf(c) = N.

... a. b.

. c... c1.. c2.. c3

TheoremSuppose that f is con nuouson the closed interval [a, b]

and let N be any numberbetween f(a) and f(b), wheref(a) ̸= f(b). Then there existsa number c in (a, b) such thatf(c) = N.

. a. b.

. c... c1.. c2.. c3

TheoremSuppose that f is con nuouson the closed interval [a, b]

and let N be any numberbetween f(a) and f(b), wheref(a) ̸= f(b). Then there existsa number c in (a, b) such thatf(c) = N.

... a. b.

. c... c1.. c2.. c3

TheoremSuppose that f is con nuouson the closed interval [a, b]and let N be any numberbetween f(a) and f(b), wheref(a) ̸= f(b).

Then there existsa number c in (a, b) such thatf(c) = N.

... a. b.

. c... c1.. c2.. c3

TheoremSuppose that f is con nuouson the closed interval [a, b]and let N be any numberbetween f(a) and f(b), wheref(a) ̸= f(b). Then there existsa number c in (a, b) such thatf(c) = N. .. x.

... a. b.

.. c1.. c2.. c3

... a. b.

. c... c1.. c2.. c3

... a. b.

.. c1.. c2.. c3

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.

S ll, it can be used in itera on or in conjunc on with othertheorems to answer these ques ons.

Using the IVT to find zeroes

Example

Let f(x) = x3 − x− 1. Show that there is a zero for f on the interval[1, 2].

Solu onf(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 bisec ons.

Example

Finding a zero by bisection

x f(x)

1 − 11.25 − 0.296875

1.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

(More careful analysis yields1.32472.)

........

x f(x)1 − 1

1.25 − 0.2968751.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

......

x f(x)1 − 1

1.25 − 0.2968751.3125 − 0.05151371.375 0.224609

1.5 0.875

x f(x)1 − 1

1.25 − 0.2968751.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

x f(x)1 − 1

1.25 − 0.296875

1.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

......

x f(x)1 − 1

1.25 − 0.296875

1.3125 − 0.0515137

1.375 0.2246091.5 0.8752 5

.......

x f(x)1 − 1

1.25 − 0.2968751.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

........

x f(x)1 − 1

1.25 − 0.2968751.3125 − 0.05151371.375 0.224609

1.5 0.8752 5

........

Using the IVT to assert existenceof numbers

Example

Suppose we are unaware of the square root func on and that it’scon nuous. Prove that the square root of two exists.

Proof.Let f(x) = x2, a con nuous func on 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.

Example

Proof.Let f(x) = x2, a con nuous func on 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.

Example

Proof.Let f(x) = x2, a con nuous func on 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.

Outline

Con nuity

Back to the QuestionsTrue 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

AnswerThe answer is TRUE.

I Let h(t) be height, which varies con nuously over me.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 2AnswerThe answer is TRUE.

I Let h(t) be height in inches and w(t) be weight in pounds, bothvarying con nuously over me.

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

AnswerThe answer is TRUE.

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

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!

SummaryWhat have we learned today?

I Defini on: a func on is con nuous at a point if the limit of thefunc on at that point agrees with the value of the func on atthat point.

I We o en make a fundamental assump on that func ons wemeet in nature are con nuous.

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

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