Lesson 6: Limits Involving Infinity

. . . . . .

Section1.6LimitsinvolvingInfinity

V63.0121.006/016, CalculusI

February3, 2010

Announcements

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

. . . . . .

Recallthedefinitionoflimit

DefinitionWewrite

limx→a

f(x) = L

andsay

“thelimitof f(x), as x approaches a, equals L”

ifwecanmakethevaluesof f(x) arbitrarilycloseto L (asclosetoL aswelike)bytaking x tobesufficientlycloseto a (oneithersideof a)butnotequalto a.

. . . . . .

Recalltheunboundednessproblem

Recallwhy limx→0+

1xdoesn’texist.

. .x

.y

..L?

Nomatterhowthinwedrawthestriptotherightof x = 0, wecannot“capture”thegraphinsidethebox.

. . . . . .


Recallwhy limx→0+

1xdoesn’texist.

. .x

.y

..L?


. . . . . .


Recallwhy limx→0+

1xdoesn’texist.

. .x

.y

..L?


. . . . . .


Recallwhy limx→0+

1xdoesn’texist.

. .x

.y

..L?


. . . . . .

Outline

InfiniteLimitsVerticalAsymptotesInfiniteLimitsweKnowLimit“Laws”withInfiniteLimitsIndeterminateLimitforms

Limitsat ∞AlgebraicratesofgrowthRationalizingtogetalimit

. . . . . .

InfiniteLimits

DefinitionThenotation

limx→a

f(x) = ∞

meansthatvaluesof f(x) canbemadearbitrarily large (aslarge asweplease)bytakingx sufficientlycloseto a butnotequalto a.

I “Large”takestheplaceof“closeto L”.

. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

InfiniteLimits


limx→a

f(x) = ∞



. .x

.y

. . . . . .

NegativeInfinity


limx→a

f(x) = −∞

meansthatthevaluesof f(x) canbemadearbitrarilylargenegative (aslargeasweplease)bytaking x sufficientlycloseto abutnotequalto a.

I Wecallanumber large or small basedonitsabsolutevalue.So −1, 000, 000 isalarge(negative)number.

. . . . . .

NegativeInfinity


limx→a

f(x) = −∞

meansthatthevaluesof f(x) canbemadearbitrarilylargenegative (aslargeasweplease)bytaking x sufficientlycloseto abutnotequalto a.

I Wecallanumber large or small basedonitsabsolutevalue.So −1, 000, 000 isalarge(negative)number.

. . . . . .

VerticalAsymptotes

DefinitionTheline x = a iscalleda verticalasymptote ofthecurve y = f(x)ifatleastoneofthefollowingistrue:

I limx→a

f(x) = ∞

I limx→a+

f(x) = ∞

I limx→a−

f(x) = ∞

I limx→a

f(x) = −∞

I limx→a+

f(x) = −∞

I limx→a−

f(x) = −∞

. . . . . .

InfiniteLimitsweKnow

I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

. .x

.y

.

.

.

.

.

.

.

.

.

.

.

.

. . . . . .


I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

. .x

.y

.

.

.

.

.

.

.

.

.

.

.

.

. . . . . .


I limx→0+

1x= ∞

I limx→0−

1x= −∞

I limx→0

1x2

= ∞

. .x

.y

.

.

.

.

.

.

.

.

.

.

.

.

. . . . . .

Findinglimitsattroublespots

ExampleLet

f(x) =x2 + 2

x2 − 3x+ 2

Find limx→a−

f(x) and limx→a+

f(x) foreach a atwhich f isnot

continuous.

SolutionThedenominatorfactorsas (x− 1)(x− 2). Wecanrecordthesignsofthefactorsonthenumberline.

. . . . . .

Findinglimitsattroublespots

ExampleLet

f(x) =x2 + 2

x2 − 3x+ 2

Find limx→a−

f(x) and limx→a+

f(x) foreach a atwhich f isnot

continuous.

SolutionThedenominatorfactorsas (x− 1)(x− 2). Wecanrecordthesignsofthefactorsonthenumberline.

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+

.+∞ .−∞ .− .−∞ .+∞ .+

So

limx→1−

f(x) = +∞ limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞

.−∞ .− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞

limx→2−

f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞

.− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞

limx→2−

f(x) = −∞

limx→1+

f(x) = −∞

limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .−

.−∞ .+∞ .+

Solim

x→1−f(x) = +∞

limx→2−

f(x) = −∞

limx→1+

f(x) = −∞

limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞

.+∞ .+

Solim

x→1−f(x) = +∞ lim

x→2−f(x) = −∞

limx→1+

f(x) = −∞

limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞

.+

Solim

x→1−f(x) = +∞ lim

x→2−f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞

.+

Solim

x→1−f(x) = +∞ lim

x→2−f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

Usethenumberline

. .(x− 1).− .

.1

.0 .+

.(x− 2).− .

.2

.0 .+

.(x2 + 2).+

.f(x)..1

..2

.+ .+∞ .−∞ .− .−∞ .+∞ .+

Solim

x→1−f(x) = +∞ lim

x→2−f(x) = −∞

limx→1+

f(x) = −∞ limx→2+

f(x) = +∞

. . . . . .

InEnglish, now

Toexplainthelimit, youcansay:“As x → 1−, thenumeratorapproaches 3, andthedenominatorapproaches 0 whileremainingpositive. Sothelimitis +∞.”

. . . . . .

Thegraphsofar

. .x

.y

..−1

..1

..2

..3

. . . . . .

Thegraphsofar

. .x

.y

..−1

..1

..2

..3

. . . . . .

Thegraphsofar

. .x

.y

..−1

..1

..2

..3

. . . . . .

Thegraphsofar

. .x

.y

..−1

..1

..2

..3

. . . . . .

Thegraphsofar

. .x

.y

..−1

..1

..2

..3

. . . . . .

LimitLaws(?) withinfinitelimits

I If limx→a

f(x) = ∞ and limx→a

g(x) = ∞, then limx→a

(f(x) + g(x)) = ∞.

Thatis,

..∞+∞ = ∞

I If limx→a

f(x) = −∞ and limx→a

g(x) = −∞, then

limx→a

(f(x) + g(x)) = −∞. Thatis,

..−∞−∞ = −∞

. . . . . .

RulesofThumb withinfinitelimits

I If limx→a

f(x) = ∞ and limx→a

g(x) = ∞, then limx→a

(f(x) + g(x)) = ∞.

Thatis,

..∞+∞ = ∞

I If limx→a

f(x) = −∞ and limx→a

g(x) = −∞, then

limx→a

(f(x) + g(x)) = −∞. Thatis,

..−∞−∞ = −∞

. . . . . .

RulesofThumbwithinfinitelimits

I If limx→a

f(x) = L and limx→a

g(x) = ±∞, then

limx→a

(f(x) + g(x)) = ±∞. Thatis,

..L+∞ = ∞L−∞ = −∞

. . . . . .

RulesofThumbwithinfinitelimitsKids, don’ttrythisathome!

I Theproductofafinitelimitandaninfinitelimitisinfinite ifthefinitelimitisnot0.

..L · ∞ =

{∞ if L > 0

−∞ if L < 0.

..L · (−∞) =

{−∞ if L > 0

∞ if L < 0.

. . . . . .

MultiplyinginfinitelimitsKids, don’ttrythisathome!

I Theproductoftwoinfinitelimitsisinfinite.

..

∞ ·∞ = ∞∞ · (−∞) = −∞

(−∞) · (−∞) = ∞

. . . . . .

DividingbyInfinityKids, don’ttrythisathome!

I Thequotientofafinitelimitbyaninfinitelimitiszero:

..L∞

= 0

. . . . . .

Dividingbyzeroisstillnotallowed

..

10= ∞

Thereareexamplesofsuchlimitformswherethelimitis ∞, −∞,undecidedbetweenthetwo, ortrulyneither.

. . . . . .

IndeterminateLimitforms

LimitsoftheformL0are indeterminate. Thereisnorulefor

evaluatingsuchaform; thelimitmustbeexaminedmoreclosely.Considerthese:

limx→0

1x2

= ∞ limx→0

−1x2

= −∞

limx→0+

1x= ∞ lim

x→0−

1x= −∞

Worst, limx→0

1x sin(1/x)

isoftheformL0, butthelimitdoesnot

exist, evenintheleft-orright-handsense. Thereareinfinitelymanyverticalasymptotesarbitrarilycloseto0!

. . . . . .

IndeterminateLimitforms

Limitsoftheform 0 · ∞ and ∞−∞ arealsoindeterminate.

Example

I Thelimit limx→0+

sin x · 1xisoftheform 0 · ∞, buttheansweris

1.


sin2 x · 1xisoftheform 0 ·∞, buttheansweris

0.


sin x · 1x2

isoftheform 0 · ∞, buttheansweris∞.

Limitsofindeterminateformsmayormaynot“exist.” Itwilldependonthecontext.

. . . . . .

IndeterminateformsarelikeTugOfWar

Whichsidewinsdependsonwhichsideisstronger.

. . . . . .

Outline

InfiniteLimitsVerticalAsymptotesInfiniteLimitsweKnowLimit“Laws”withInfiniteLimitsIndeterminateLimitforms

Limitsat ∞AlgebraicratesofgrowthRationalizingtogetalimit

. . . . . .

DefinitionLet f beafunctiondefinedonsomeinterval (a,∞). Then

limx→∞

f(x) = L

meansthatthevaluesof f(x) canbemadeascloseto L aswelike, bytaking x sufficientlylarge.

DefinitionTheline y = L isacalleda horizontalasymptote ofthecurvey = f(x) ifeither

limx→∞

f(x) = L or limx→−∞

f(x) = L.

y = L isa horizontal line!

. . . . . .


limx→∞

f(x) = L



limx→∞


f(x) = L.


. . . . . .


limx→∞

f(x) = L



limx→∞


f(x) = L.


. . . . . .

Basiclimitsatinfinity

TheoremLet n beapositiveinteger. Then

I limx→∞

1xn

= 0

I limx→−∞

1xn

= 0

. . . . . .

Usingthelimitlawstocomputelimitsat ∞

ExampleFind

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

ifitexists.

A doesnotexist

B 1/2

C 0

D ∞

. . . . . .

Usingthelimitlawstocomputelimitsat ∞

ExampleFind

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

ifitexists.

A doesnotexist

B 1/2

C 0

D ∞

. . . . . .

SolutionFactoroutthelargestpowerof x fromthenumeratoranddenominator. Wehave

2x3 + 3x+ 14x3 + 5x2 + 7

=x3(2+ 3/x2 + 1/x3)

x3(4+ 5/x + 7/x3)

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

= limx→∞

2+ 3/x2 + 1/x3

4+ 5/x + 7/x3

=2+ 0+ 04+ 0+ 0

=12

UpshotWhenfindinglimitsofalgebraicexpressionsatinfinity, lookatthe highestdegreeterms.

. . . . . .

SolutionFactoroutthelargestpowerof x fromthenumeratoranddenominator. Wehave

2x3 + 3x+ 14x3 + 5x2 + 7

=x3(2+ 3/x2 + 1/x3)

x3(4+ 5/x + 7/x3)

limx→∞

2x3 + 3x+ 14x3 + 5x2 + 7

= limx→∞

2+ 3/x2 + 1/x3

4+ 5/x + 7/x3

=2+ 0+ 04+ 0+ 0

=12

UpshotWhenfindinglimitsofalgebraicexpressionsatinfinity, lookatthe highestdegreeterms.

. . . . . .

AnotherExample

ExampleFind lim

x→∞

xx2 + 1

AnswerThelimitis 0.

. .x

.y

Noticethatthegraphdoescrosstheasymptote, whichcontradictsoneoftheheuristicdefinitionsofasymptote.

. . . . . .

AnotherExample

ExampleFind lim

x→∞

xx2 + 1

AnswerThelimitis 0.

. .x

.y


. . . . . .

SolutionAgain, factoroutthelargestpowerof x fromthenumeratoranddenominator. Wehave

xx2 + 1

=x(1)

x2(1+ 1/x2)=

1x· 11+ 1/x2

limx→∞

xx2 + 1

= limx→∞

1x

11+ 1/x2

= limx→∞

1x· limx→∞

11+ 1/x2

= 0 · 11+ 0

= 0.

RemarkHadthehigherpowerbeeninthenumerator, thelimitwouldhavebeen ∞.

. . . . . .

AnotherExample

ExampleFind lim

x→∞

xx2 + 1

AnswerThelimitis 0.

. .x

.y


. . . . . .

AnotherExample

ExampleFind lim

x→∞

xx2 + 1

AnswerThelimitis 0.

. .x

.y


. . . . . .

SolutionAgain, factoroutthelargestpowerof x fromthenumeratoranddenominator. Wehave

xx2 + 1

=x(1)

x2(1+ 1/x2)=

1x· 11+ 1/x2

limx→∞

xx2 + 1

= limx→∞

1x

11+ 1/x2

= limx→∞

1x· limx→∞

11+ 1/x2

= 0 · 11+ 0

= 0.

RemarkHadthehigherpowerbeeninthenumerator, thelimitwouldhavebeen ∞.

. . . . . .

AnotherExample

ExampleFind

limx→∞

√3x4 + 7x2 + 3

..√3x4 + 7 ∼

√3x4 =

√3x2

AnswerThelimitis

√3.

. . . . . .

AnotherExample

ExampleFind

limx→∞

√3x4 + 7x2 + 3

..√3x4 + 7 ∼

√3x4 =

√3x2

AnswerThelimitis

√3.

. . . . . .

Solution

limx→∞

√3x4 + 7x2 + 3

= limx→∞

√x4(3+ 7/x4)

x2(1+ 3/x2)

= limx→∞

x2√

(3+ 7/x4)

x2(1+ 3/x2)

= limx→∞

√(3+ 7/x4)

1+ 3/x2

=

√3+ 01+ 0

=√3.

. . . . . .

Rationalizingtogetalimit

ExampleCompute lim

x→∞

(√4x2 + 17− 2x

).

SolutionThislimitisoftheform ∞−∞, whichwecannotuse. Sowerationalizethenumerator(thedenominatoris 1)togetanexpressionthatwecanusethelimitlawson.

limx→∞

(√4x2 + 17− 2x

)= lim

x→∞

(√4x2 + 17− 2x

)·√4x2 + 17+ 2x√4x2 + 17+ 2x

= limx→∞

(4x2 + 17)− 4x2√4x2 + 17+ 2x

= limx→∞

17√4x2 + 17+ 2x

= 0

. . . . . .

Rationalizingtogetalimit

ExampleCompute lim

x→∞

(√4x2 + 17− 2x

).

SolutionThislimitisoftheform ∞−∞, whichwecannotuse. Sowerationalizethenumerator(thedenominatoris 1)togetanexpressionthatwecanusethelimitlawson.

limx→∞

(√4x2 + 17− 2x

)= lim

x→∞

(√4x2 + 17− 2x

)·√4x2 + 17+ 2x√4x2 + 17+ 2x

= limx→∞

(4x2 + 17)− 4x2√4x2 + 17+ 2x

= limx→∞

17√4x2 + 17+ 2x

= 0

. . . . . .

Kickitupanotch

ExampleCompute lim

x→∞

(√4x2 + 17x− 2x

).

SolutionSametrick, differentanswer:

limx→∞

(√4x2 + 17x− 2x

)= lim

x→∞

(√4x2 + 17x− 2x

)·√4x2 + 17+ 2x√4x2 + 17x+ 2x

= limx→∞

(4x2 + 17x)− 4x2√4x2 + 17x+ 2x

= limx→∞

17x√4x2 + 17x+ 2x

= limx→∞

17√4+ 17/x+ 2

=174

. . . . . .

Kickitupanotch

ExampleCompute lim

x→∞

(√4x2 + 17x− 2x

).

SolutionSametrick, differentanswer:

limx→∞

(√4x2 + 17x− 2x

)= lim

x→∞

(√4x2 + 17x− 2x

)·√4x2 + 17+ 2x√4x2 + 17x+ 2x

= limx→∞

(4x2 + 17x)− 4x2√4x2 + 17x+ 2x

= limx→∞

17x√4x2 + 17x+ 2x

= limx→∞

17√4+ 17/x+ 2

=174

. . . . . .

Summary

I Infinityisamorecomplicatedconceptthanasinglenumber.Therearerulesofthumb, buttherearealsoexceptions.

I Takeatwo-prongedapproachtolimitsinvolvinginfinity:I Lookattheexpressiontoguessthelimit.I Uselimitrulesandalgebratoverifyit.

Lesson 6: Limits Involving Infinity

Education