Lesson 6: Limits Involving Infinity
70
. . . . . . Section 1.6 Limits involving Infinity V63.0121.027, Calculus I September 22, 2009 Announcements I Quiz 1 this week (§§1.1–1.3) I Written Assignment 2 due Thursday
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Infinity is a dangerous place where the rules of arithmetic break down. But it is a useful concept and study both infinite limits and limits at infinity.
Transcript of Lesson 6: Limits Involving Infinity
. . . . . .
Section1.6LimitsinvolvingInfinity
V63.0121.027, CalculusI
September22, 2009
Announcements
I Quiz1thisweek(§§1.1–1.3)I WrittenAssignment2dueThursday
. . . . . .
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.
. . . . . .
Recalltheunboundednessproblem
Recallwhy limx→0+
1xdoesn’texist.
. .x
.y
..L?
Nomatterhowthinwedrawthestriptotherightof x = 0, wecannot“capture”thegraphinsidethebox.
. . . . . .
Recalltheunboundednessproblem
Recallwhy limx→0+
1xdoesn’texist.
. .x
.y
..L?
Nomatterhowthinwedrawthestriptotherightof x = 0, wecannot“capture”thegraphinsidethebox.
. . . . . .
Recalltheunboundednessproblem
Recallwhy limx→0+
1xdoesn’texist.
. .x
.y
..L?
Nomatterhowthinwedrawthestriptotherightof x = 0, wecannot“capture”thegraphinsidethebox.
. . . . . .
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
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
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
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
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
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
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
DefinitionThenotation
limx→a
f(x) = ∞
meansthatvaluesof f(x) canbemadearbitrarily large (aslarge asweplease)bytakingx sufficientlycloseto a butnotequalto a.
I “Large”takestheplaceof“closeto L”.
. .x
.y
. . . . . .
NegativeInfinity
DefinitionThenotation
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
DefinitionThenotation
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
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . .
InfiniteLimitsweKnow
I limx→0+
1x
= ∞
I limx→0−
1x
= −∞
I limx→0
1x2
= ∞
. .x
.y
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . .
InfiniteLimitsweKnow
I limx→0+
1x
= ∞
I limx→0−
1x
= −∞
I limx→0
1x2
= ∞
. .x
.y
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . .
Findinglimitsattroublespots
ExampleLet
f(t) =t2 + 2
t2 − 3t + 2
Find limt→a−
f(t) and limt→a+
f(t) foreach a atwhich f isnotcontinuous.
SolutionThedenominatorfactorsas (t− 1)(t− 2). Wecanrecordthesignsofthefactorsonthenumberline.
. . . . . .
Findinglimitsattroublespots
ExampleLet
f(t) =t2 + 2
t2 − 3t + 2
Find limt→a−
f(t) and limt→a+
f(t) foreach a atwhich f isnotcontinuous.
SolutionThedenominatorfactorsas (t− 1)(t− 2). Wecanrecordthesignsofthefactorsonthenumberline.
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+
.±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞
.− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .−
.∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞
.+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
Usethenumberline
. .(t− 1).− .
.1
.0 .+
.(t− 2).− .
.2
.0 .+
.(t2 + 2).+
.f(t)..1
..2
.+ .±∞ .− .∓∞ .+
Solimx→1−
f(x) = +∞ limx→2−
f(x) = −∞
limx→1+
f(x) = −∞ limx→2+
f(x) = +∞
. . . . . .
InEnglish, now
Toexplainthelimit, youcansay:“As t → 1−, thenumeratorapproaches 2, 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
= ∞ limx→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.
I Thelimit limx→0+
sin2 x · 1xisoftheform 0 ·∞, buttheansweris
0.
I Thelimit limx→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!
. . . . . .
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!
. . . . . .
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!
. . . . . .
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
Noticethatthegraphdoescrosstheasymptote, whichcontradictsoneoftheheuristicdefinitionsofasymptote.
. . . . . .
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
Noticethatthegraphdoescrosstheasymptote, whichcontradictsoneoftheheuristicdefinitionsofasymptote.
. . . . . .
AnotherExample
ExampleFind lim
x→∞
xx2 + 1
AnswerThelimitis 0.
. .x
.y
Noticethatthegraphdoescrosstheasymptote, whichcontradictsoneoftheheuristicdefinitionsofasymptote.
. . . . . .
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
. . . . . .
Summary
I Infinityisamorecomplicatedconceptthanasinglenumber.Therearerulesofthumb, buttherearealsoexceptions.
I Takeatwo-prongedapproachtolimitsinvolvinginfinity:I Lookattheexpressiontoguessthelimit.I Uselimitrulesandalgebratoverifyit.
