Download - Lesson 5: Continuity

. . . . . .

Section1.5Continuity

V63.0121.006/016, CalculusI

February2, 2010

Announcements

I OfficeHours: M,W 1:30–2:30, R 9–10(CIWW 726)I WrittenAssignment#2dueThursdayI FirstQuiz: FridayFebruary12inrecitation(§§1.1–1.4)

. . . . . .

Hatsumon

Herearesomediscussionquestionstostart.

TrueorFalseAtsomepointinyourlifeyouwereexactlythreefeettall.

TrueorFalseAtsomepointinyourlifeyourheight(ininches)wasequaltoyourweight(inpounds).

TrueorFalseRightnowthereareapairofpointsonoppositesidesoftheworldmeasuringtheexactsametemperature.

. . . . . .

Outline

Continuity

TheIntermediateValueTheorem

BacktotheQuestions

. . . . . .

Recall: DirectSubstitutionProperty

Theorem(TheDirectSubstitutionProperty)If f isapolynomialorarationalfunctionand a isinthedomainoff, then

limx→a

f(x) = f(a)

Thispropertyissousefulit’sworthnaming.

. . . . . .

DefinitionofContinuity

Definition

I Let f beafunctiondefinednear a. Wesaythat f is continuousat aif

limx→a

f(x) = f(a).

I A function f iscontinuous ifitiscontinuousateverypointinitsdomain.

. .x

.y

.

.a

.f(a)

. . . . . .

Scholium

DefinitionLet f beafunctiondefinednear a. Wesaythat f is continuousata if

limx→a

f(x) = f(a).

Therearethreeimportantpartstothisdefinition.I Thefunctionhastohavealimitat a,I thefunctionhastohaveavalueat a,I andthesevalueshavetoagree.

. . . . . .

FreeTheorems

Theorem

(a) Anypolynomialiscontinuouseverywhere; thatis, itiscontinuouson R = (−∞,∞).

(b) Anyrationalfunctioniscontinuouswhereveritisdefined;thatis, itiscontinuousonitsdomain.

. . . . . .

Showingafunctioniscontinuous

ExampleLet f(x) =

√4x+ 1. Showthat f iscontinuousat 2.

SolutionWewanttoshowthat lim

x→2f(x) = f(2). Wehave

limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).

Eachstepcomesfromthelimitlaws.

QuestionAtwhichotherpointsis f continuous?

AnswerThefunction f iscontinuouson (−1/4,∞). Itis rightcontinuousat −1/4 since lim

x→−1/4+f(x) = f(−1/4).

. . . . . .

Atwhichotherpoints?

Forreference: 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 iscontinuousat a.

I If a = −1/4, then 4x+ 1 < 0 totheleftof a, whichmeans√4x+ 1 isundefined. Still,

limx→a+

f(x) = limx→a+

√4x+ 1 =

√limx→a+

(4x+ 1) =√0 = 0 = f(a)

so f iscontinuousontherightat a = −1/4.

. . . . . .

Atwhichotherpoints?

Forreference: 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 iscontinuousat a.I If a = −1/4, then 4x+ 1 < 0 totheleftof a, whichmeans√

4x+ 1 isundefined. Still,

limx→a+

f(x) = limx→a+

√4x+ 1 =

√limx→a+

(4x+ 1) =√0 = 0 = f(a)

so f iscontinuousontherightat a = −1/4.

. . . . . .


ExampleLet f(x) =




limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).



AnswerThefunction f iscontinuouson (−1/4,∞).

Itis rightcontinuousat −1/4 since lim

x→−1/4+f(x) = f(−1/4).

. . . . . .


ExampleLet f(x) =




limx→a

f(x) = limx→2

√4x+ 1 =

√limx→2

(4x+ 1) =√9 = 3 = f(2).



AnswerThefunction f iscontinuouson (−1/4,∞). Itis rightcontinuousat −1/4 since lim

x→−1/4+f(x) = f(−1/4).

. . . . . .

TheLimitLawsgiveContinuityLaws

TheoremIf f(x) and g(x) arecontinuousat a and c isaconstant, thenthefollowingfunctionsarealsocontinuousat a:

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

Ifg(x) (if g(a) ̸= 0)

. . . . . .

Whyasumofcontinuousfunctionsiscontinuous

Wewanttoshowthat

limx→a

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

Wejustfollowournose:

limx→a

(f+ g)(x) = limx→a

[f(x) + g(x)] (defof f+ g)

= limx→a

f(x) + limx→a

g(x) (iftheselimitsexist)

= f(a) + g(a) (theydo; f and g arects.)

= (f+ g)(a) (defof f+ g again)

. . . . . .

Trigonometricfunctionsarecontinuous

I sin and cos arecontinuouson R.

I tan =sincos

and sec =1cos

arecontinuousontheirdomain, whichisR \

{ π

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin

arecontinuousontheirdomain, whichisR \ { kπ | k ∈ Z }.

..sin

.cos

.tan .sec

.cot .csc

. . . . . .



I tan =sincos

and sec =1cos


{ π

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin


..sin

.cos

.tan

.sec

.cot .csc

. . . . . .



I tan =sincos

and sec =1cos


{ π

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin


..sin

.cos

.tan .sec

.cot .csc

. . . . . .



I tan =sincos

and sec =1cos


{ π

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin


..sin

.cos

.tan .sec

.cot

.csc

. . . . . .



I tan =sincos

and sec =1cos


{ π

2+ kπ

∣∣∣ k ∈ Z}.

I cot =cossin

and csc =1sin


..sin

.cos

.tan .sec

.cot .csc

. . . . . .

ExponentialandLogarithmicfunctionsarecontinuous

Foranybase a > 1,

I thefunction x 7→ ax iscontinuouson R

I thefunction loga iscontinuousonitsdomain: (0,∞)

I Inparticular ex andln = loge arecontinuousontheirdomains

.

.ax

.loga x

. . . . . .

InversetrigonometricfunctionsaremostlycontinuousI sin−1 and cos−1 arecontinuouson (−1,1), leftcontinuousat 1, andrightcontinuousat −1.

I sec−1 and csc−1 arecontinuouson (−∞,−1) ∪ (1,∞), leftcontinuousat −1, andrightcontinuousat 1.

I tan−1 and cot−1 arecontinuouson R.

.

.−π

.−π/2

.π/2

.π

.

.

.sin−1

.

.

.cos−1.sec−1

.

..csc−1

.

.

.tan−1

.cot−1

. . . . . .




.

.−π

.−π/2

.π/2

.π

.

.

.sin−1.

.

.cos−1

.sec−1

.

..csc−1

.

.

.tan−1

.cot−1

. . . . . .




.

.−π

.−π/2

.π/2

.π

.

.

.sin−1.

.

.cos−1.sec−1

.

.

.csc−1

.

.

.tan−1

.cot−1

. . . . . .




.

.−π

.−π/2

.π/2

.π

.

.

.sin−1.

.

.cos−1.sec−1

.

..csc−1

.

.

.tan−1

.cot−1

. . . . . .

Whatcouldgowrong?

Inwhatwayscouldafunction f failtobecontinuousatapoint a?Lookagainatthedefinition:

limx→a

f(x) = f(a)

. . . . . .

ContinuityFAIL

:Thelimitdoesnotexist

ExampleLet

f(x) =

{x2 if 0 ≤ x ≤ 1

2x if 1 < x ≤ 2

Atwhichpointsis f continuous?

SolutionAtanypoint a in [0, 2] besides 1, lim

x→af(x) = f(a) because f is

representedbyapolynomialnear a, andpolynomialshavethedirectsubstitutionproperty. However,

limx→1−

f(x) = limx→1−

x2 = 12 = 1

limx→1+

f(x) = limx→1+

2x = 2(1) = 2

So f hasnolimitat 1. Therefore f isnotcontinuousat 1.

. . . . . .

ContinuityFAIL:Thelimitdoesnotexist

ExampleLet

f(x) =

{x2 if 0 ≤ x ≤ 1

2x if 1 < x ≤ 2


SolutionAtanypoint a in [0, 2] besides 1, lim

x→af(x) = f(a) because f is

representedbyapolynomialnear a, andpolynomialshavethedirectsubstitutionproperty. However,

limx→1−

f(x) = limx→1−

x2 = 12 = 1

limx→1+

f(x) = limx→1+

2x = 2(1) = 2

So f hasnolimitat 1. Therefore f isnotcontinuousat 1.

. . . . . .

GraphicalIllustrationofPitfall#1

. .x

.y

..−1

..1

..2

..−1

..1

..2

..3

..4

.

.

.

.

. . . . . .

ContinuityFAIL

:Thefunctionhasnovalue

ExampleLet

f(x) =x2 + 2x+ 1

x+ 1


SolutionBecause f isrational, itiscontinuousonitswholedomain. Notethat −1 isnotinthedomainof f, so f isnotcontinuousthere.

. . . . . .

ContinuityFAIL:Thefunctionhasnovalue

ExampleLet

f(x) =x2 + 2x+ 1

x+ 1


SolutionBecause f isrational, itiscontinuousonitswholedomain. Notethat −1 isnotinthedomainof f, so f isnotcontinuousthere.

. . . . . .


. .x

.y

...−1

. .1

f cannotbecontinuouswhereithasnovalue.

. . . . . .

ContinuityFAIL

:functionvalue ̸= limit

ExampleLet

f(x) =

{7 if x ̸= 1

π if x = 1


Solutionf isnotcontinuousat 1 because f(1) = π but lim

x→1f(x) = 7.

. . . . . .

ContinuityFAIL:functionvalue ̸= limit

ExampleLet

f(x) =

{7 if x ̸= 1

π if x = 1


Solutionf isnotcontinuousat 1 because f(1) = π but lim

x→1f(x) = 7.

. . . . . .


. .x

.y

..π

..7

..1

.

.

. . . . . .

Specialtypesofdiscontinuites

removablediscontinuity Thelimit limx→a

f(x) exists, but f isnot

definedat a oritsvalueat a isnotequaltothelimitat a.

Byre-defining f(a) = limx→a

f(x), f canbemade

continuousat a

jumpdiscontinuity Thelimits limx→a−

f(x) and limx→a+

f(x) exist, but

aredifferent.

Thefunctioncannotbemadecontinuousbychangingasinglevalue.

. . . . . .

Graphicalrepresentationsofdiscontinuities

. .x

.y

..π

..7

..1

.

.

..Presto! continuous!

removable

. .x

.y

..1

..1

..2

.

.

.

.

. .continuous?.

. .continuous?

. .continuous?

jump

. . . . . .

Graphicalrepresentationsofdiscontinuities

. .x

.y

..π

..7

..1

.

.

..Presto! continuous!

removable

. .x

.y

..1

..1

..2

.

.

.

.

. .continuous?

.

. .continuous?

. .continuous?

jump

. . . . . .




definedat a oritsvalueat a isnotequaltothelimitat a. Byre-defining f(a) = lim

x→af(x), f canbemade

continuousat a


f(x) and limx→a+

f(x) exist, but

aredifferent.

Thefunctioncannotbemadecontinuousbychangingasinglevalue.

. . . . . .




definedat a oritsvalueat a isnotequaltothelimitat a. Byre-defining f(a) = lim

x→af(x), f canbemade

continuousat a


f(x) and limx→a+

f(x) exist, but

aredifferent. Thefunctioncannotbemadecontinuousbychangingasinglevalue.

. . . . . .

Thegreatestintegerfunction

[[x]] isthegreatestinteger ≤ x.

x [[x]]0 01 1

1.5 11.9 12.1 2

−0.5 −1−0.9 −1−1.1 −2

. .x

.y

..−2

..−2

..−1

..−1

..1

..1

..2

..2

..3

..3

. .

. .

. .

. .

. ..y = [[x]]

Thisfunctionhasajumpdiscontinuityateachinteger.

. . . . . .

Outline

Continuity


BacktotheQuestions

. . . . . .

A BigTimeTheorem

Theorem(TheIntermediateValueTheorem)Supposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.

. . . . . .

IllustratingtheIVT

Supposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.

. .x

.f(x)

.

.

.a .b

.f(a)

.f(b)

.N

.c

..

.c1

.

.c2

.

.c3

. . . . . .

IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b]

andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.

. .x

.f(x)

.

.

.a .b

.f(a)

.f(b)

.N

.c

..

.c1

.

.c2

.

.c3

. . . . . .

IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b).

Thenthereexistsanumber c in (a,b) suchthat f(c) = N.

. .x

.f(x)

.

.

.a .b

.f(a)

.f(b)

.N

.c

..

.c1

.

.c2

.

.c3

. . . . . .

IllustratingtheIVTSupposethat f iscontinuousontheclosedinterval [a,b] andlet Nbeanynumberbetween f(a) and f(b), where f(a) ̸= f(b). Thenthereexistsanumber c in (a,b) suchthat f(c) = N.

. .x

.f(x)

.

.

.a .b

.f(a)

.f(b)

.N

.c

.

.

.c1

.

.c2

.

.c3

. . . . . .


. .x

.f(x)

.

.

.a .b

.f(a)

.f(b)

.N

.c

..

.c1

.

.c2

.

.c3

. . . . . .


. .x

.f(x)

.

.

.a .b

.f(a)

.f(b)

.N

.c

.

.

.c1

.

.c2

.

.c3

. . . . . .

WhattheIVT doesnotsay

TheIntermediateValueTheoremisan“existence”theorem.I Itdoesnotsayhowmanysuch c exist.I Italsodoesnotsayhowtofind c.

Still, itcanbeusediniterationorinconjunctionwithothertheoremstoanswerthesequestions.

. . . . . .

UsingtheIVT

ExampleSupposeweareunawareofthesquarerootfunctionandthatit’scontinuous. Provethatthesquarerootoftwoexists.

Proof.Let f(x) = x2, acontinuousfunctionon [1, 2]. Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat

f(c) = c2 = 2.

Infact, wecan“narrowin”onthesquarerootof 2 by themethodofbisections.

. . . . . .

UsingtheIVT


Proof.Let f(x) = x2, acontinuousfunctionon [1, 2].

Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat

f(c) = c2 = 2.


. . . . . .

UsingtheIVT


Proof.Let f(x) = x2, acontinuousfunctionon [1, 2]. Note f(1) = 1 andf(2) = 4. Since 2 isbetween 1 and 4, thereexistsapoint c in(1, 2) suchthat

f(c) = c2 = 2.


. . . . . .

Finding√2 bybisections

.

.x .f(x) = x2

..1 .1

..2 .4

..1.5 .2.25

..1.25 .1.5625

..1.375 .1.890625

..1.4375 .2.06640625

. . . . . .

UsingtheIVT

ExampleLet f(x) = x3 − x− 1. Showthatthereisazerofor f.

Solutionf(1) = −1 and f(2) = 5. Sothereisazerobetween 1 and 2.

(Morecarefulanalysisyields 1.32472.)

. . . . . .

UsingtheIVT

ExampleLet f(x) = x3 − x− 1. Showthatthereisazerofor f.

Solutionf(1) = −1 and f(2) = 5. Sothereisazerobetween 1 and 2.(Morecarefulanalysisyields 1.32472.)

. . . . . .

Outline

Continuity


BacktotheQuestions

. . . . . .

BacktotheQuestions

TrueorFalseAtonepointinyourlifeyouwereexactlythreefeettall.

TrueorFalseAtonepointinyourlifeyourheightininchesequaledyourweightinpounds.

TrueorFalseRightnowtherearetwopointsonoppositesidesoftheEarthwithexactlythesametemperature.

. . . . . .

Question1: True!

I Let h(t) beheight, whichvariescontinuouslyovertime.I Then h(birth) < 3 ft and h(now) > 3 ft.I SobytheIVT thereisapoint c in (birth, now) where

h(c) = 3.

. . . . . .

BacktotheQuestions




. . . . . .

Question2: True!

I Let h(t) beheightininchesand w(t) beweightinpounds,bothvaryingcontinuouslyovertime.

I Let f(t) = h(t)−w(t).I Formostofus(callyourmom), f(birth) > 0 and f(now) < 0.I SobytheIVT thereisapoint c in (birth, now) where

f(c) = 0.I Inotherwords,

h(c)−w(c) = 0 ⇐⇒ h(c) = w(c).

. . . . . .

BacktotheQuestions




. . . . . .

Question3

I Let T(θ) bethetemperatureatthepointontheequatoratlongitude θ.

I Howcanyouexpressthestatementthatthetemperatureonoppositesidesisthesame?

I Howcanyouensurethisistrue?

. . . . . .

Question3: 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 Sosomewherebetween 0 and 180 thereisapoint θ wheref(θ) = 0!

. . . . . .

Whathavewelearnedtoday?

I Definition: afunctioniscontinuousatapointifthelimitofthefunctionatthatpointagreeswiththevalueofthefunctionatthatpoint.

I Weoftenmakeafundamentalassumptionthatfunctionswemeetinnaturearecontinuous.

I TheIntermediateValueTheoremisabasicpropertyofrealnumbersthatweneedandusealot.

Download - Lesson 5: Continuity

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