Lesson 6: Limits Involving Infinity
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. . . . . .
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.
. . . . . .
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(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.
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
. . . . . .
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.
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