. . . . . .
Section4.5IndeterminateFormsandL’Hôpital’s
Rule
Math1aIntroductiontoCalculus
March31, 2008
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ MidtermII:4/11inclass
..Image: Flickruser Orianonicolau
. . . . . .
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ MidtermII:4/11inclass
. . . . . .
Outline
Lasttime
IndeterminateForms
L’Hôpital’sRuleApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
TheCauchyMeanValueTheorem(Bonus)
Nexttime
. . . . . .
StrategyforOptimizationProblems
1. UnderstandtheProblem. Whatisknown? Whatisunknown? Whataretheconditions?
2. Drawadiagram.
3. IntroduceNotation.
4. Expressthe“objectivefunction” Q intermsoftheothersymbols
5. If Q isafunctionofmorethanone“decisionvariable”, usethegiveninformationtoeliminateallbutoneofthem.
6. Findtheabsolutemaximum(orminimum, dependingontheproblem)ofthefunctiononitsdomain.
. . . . . .
Outline
Lasttime
IndeterminateForms
L’Hôpital’sRuleApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
TheCauchyMeanValueTheorem(Bonus)
Nexttime
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
◮ Limitofasumisthesumofthelimits
◮ Limitofadifferenceisthedifferenceofthelimits◮ Limitofaproductistheproductofthelimits◮ Limitofaquotientisthequotientofthelimits... whoops!
Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
◮ Limitofasumisthesumofthelimits◮ Limitofadifferenceisthedifferenceofthelimits
◮ Limitofaproductistheproductofthelimits◮ Limitofaquotientisthequotientofthelimits... whoops!
Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
◮ Limitofasumisthesumofthelimits◮ Limitofadifferenceisthedifferenceofthelimits◮ Limitofaproductistheproductofthelimits
◮ Limitofaquotientisthequotientofthelimits... whoops!Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
◮ Limitofasumisthesumofthelimits◮ Limitofadifferenceisthedifferenceofthelimits◮ Limitofaproductistheproductofthelimits◮ Limitofaquotientisthequotientofthelimits... whoops!
Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Weknowdividingbyzeroisbad. Mostofthetime, ifyouhaveanumeratorwhichapproachesafinitenumberandadenominatorwhichapproacheszero, thequotientapproachessomekindofinfinity. Anexceptionwouldbesomethinglike
limx→∞
11x sin x
= limx→∞
x sec x.
whichdoesn’texist.
Evenworseisthesituationwherethenumeratoranddenominatorbothgotozero.
. . . . . .
Weknowdividingbyzeroisbad. Mostofthetime, ifyouhaveanumeratorwhichapproachesafinitenumberandadenominatorwhichapproacheszero, thequotientapproachessomekindofinfinity. Anexceptionwouldbesomethinglike
limx→∞
11x sin x
= limx→∞
x sec x.
whichdoesn’texist.Evenworseisthesituationwherethenumeratoranddenominatorbothgotozero.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experiments
◮ limx→0+
sin2 xx
= 0
◮ limx→0
xsin2 x
doesnotexist
◮ limx→0
sin2 xsin x2
= 1
◮ limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
LanguageNoteItdependsonwhatthemeaningoftheword“is”is
◮ Becarefulwiththelanguagehere. Weare not sayingthat
thelimitineachcase“is”00, andthereforenonexistent
becausethisexpressionisundefined.
◮ Thelimit isoftheform00, whichmeanswecannotevaluate
itwithourlimitlaws.
. . . . . .
IndeterminateformsarelikeTugOfWar
Whichsidewinsdependsonwhichsideisstronger.
. . . . . .
Outline
Lasttime
IndeterminateForms
L’Hôpital’sRuleApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
TheCauchyMeanValueTheorem(Bonus)
Nexttime
. . . . . .
QuestionIf f and g arelinesand f(a) = g(a) = 0, whatis
limx→a
f(x)g(x)
?
SolutionThefunctions f and g canbewrittenintheform
f(x) = m1(x− a)
g(x) = m2(x− a)
Sof(x)g(x)
=m1
m2=
f′(x)g′(x)
.
Butwhatifthefunctionsaren’tlinear? Ifonlytherewereawaytodealwithfunctionswhichwereonlyapproximatelylinear!
. . . . . .
QuestionIf f and g arelinesand f(a) = g(a) = 0, whatis
limx→a
f(x)g(x)
?
SolutionThefunctions f and g canbewrittenintheform
f(x) = m1(x− a)
g(x) = m2(x− a)
Sof(x)g(x)
=m1
m2=
f′(x)g′(x)
.
Butwhatifthefunctionsaren’tlinear? Ifonlytherewereawaytodealwithfunctionswhichwereonlyapproximatelylinear!
. . . . . .
QuestionIf f and g arelinesand f(a) = g(a) = 0, whatis
limx→a
f(x)g(x)
?
SolutionThefunctions f and g canbewrittenintheform
f(x) = m1(x− a)
g(x) = m2(x− a)
Sof(x)g(x)
=m1
m2=
f′(x)g′(x)
.
Butwhatifthefunctionsaren’tlinear? Ifonlytherewereawaytodealwithfunctionswhichwereonlyapproximatelylinear!
. . . . . .
Theorem(L’Hopital’sRule)Suppose f and g aredifferentiablefunctionsand g′(x) ̸= 0 near a(exceptpossiblyat a). Supposethat
limx→a
f(x) = 0 and limx→a
g(x) = 0
or
limx→a
f(x) = ±∞ and limx→a
g(x) = ±∞
Then
limx→a
f(x)g(x)
= limx→a
f′(x)g′(x)
,
ifthelimitontheright-handsideisfinite, ∞, or −∞.
L’Hôpital’srulealsoappliesforlimitsoftheform∞∞
.
. . . . . .
Theorem(L’Hopital’sRule)Suppose f and g aredifferentiablefunctionsand g′(x) ̸= 0 near a(exceptpossiblyat a). Supposethat
limx→a
f(x) = 0 and limx→a
g(x) = 0
or
limx→a
f(x) = ±∞ and limx→a
g(x) = ±∞
Then
limx→a
f(x)g(x)
= limx→a
f′(x)g′(x)
,
ifthelimitontheright-handsideisfinite, ∞, or −∞.
L’Hôpital’srulealsoappliesforlimitsoftheform∞∞
.
. . . . . .
MeettheMathematician: L’Hôpital
◮ wantedtobeamilitaryman, butpooreyesightforcedhimintomath
◮ didsomemathonhisown(solvedthe“brachistocroneproblem”)
◮ paidastipendtoJohannBernoulli, whoprovedthistheoremandnameditafterhim! GuillaumeFrançoisAntoine,
MarquisdeL’Hôpital(1661–1704)
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Howdoesthisaffectourexamplesabove?
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x cos x1
= 0.
Example
limx→0
sin2 xsin x2
H= lim
x→0
✓2 sin x cos x(cos x2) (✓2x)
H= lim
x→0
cos2 x− sin2 xcos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
SketchofProofofL’Hôpital’sRule
Let x beanumbercloseto a. Weknowthatf(x) − f(a)
x− a= f′(c),
forsome c ∈ (a, x); alsog(x) − g(a)
x− a= g′(d), forsome d ∈ (a, x).
Thismeansf(x)g(x)
=f′(c)g′(d)
.
ThemiracleoftheMVT isthatatweakingofitallowsustoassume c = d, sothat
f(x)g(x)
=f′(c)g′(c)
.
Thenumber c dependson x andsinceitisbetween a and x, wemusthave
limx→a
c(x) = a.
. . . . . .
BewareofRedHerrings
ExampleFind
limx→0
xcos x
SolutionThelimitofthedenominatoris 1, not 0, so L’Hôpital’sruledoesnotapply. Thelimitis 0.
. . . . . .
BewareofRedHerrings
ExampleFind
limx→0
xcos x
SolutionThelimitofthedenominatoris 1, not 0, so L’Hôpital’sruledoesnotapply. Thelimitis 0.
. . . . . .
TheoremLet r beanypositivenumber. Then
limx→∞
ex
xr= ∞.
Proof.If r isapositiveinteger, thenapplyL’Hôpital’srule r timestothefraction. Youget
limx→∞
ex
xrH= . . .
H= lim
x→∞
ex
r!= ∞.
If r isnotaninteger, let n = [[x]] and m = n + 1. Thenif x > 1,xn < xr < xm, so
ex
xn>
ex
xr>
ex
xm.
NowapplytheSqueezeTheorem.
. . . . . .
TheoremLet r beanypositivenumber. Then
limx→∞
ex
xr= ∞.
Proof.If r isapositiveinteger, thenapplyL’Hôpital’srule r timestothefraction. Youget
limx→∞
ex
xrH= . . .
H= lim
x→∞
ex
r!= ∞.
If r isnotaninteger, let n = [[x]] and m = n + 1. Thenif x > 1,xn < xr < xm, so
ex
xn>
ex
xr>
ex
xm.
NowapplytheSqueezeTheorem.
. . . . . .
TheoremLet r beanypositivenumber. Then
limx→∞
ex
xr= ∞.
Proof.If r isapositiveinteger, thenapplyL’Hôpital’srule r timestothefraction. Youget
limx→∞
ex
xrH= . . .
H= lim
x→∞
ex
r!= ∞.
If r isnotaninteger, let n = [[x]] and m = n + 1. Thenif x > 1,xn < xr < xm, so
ex
xn>
ex
xr>
ex
xm.
NowapplytheSqueezeTheorem.
. . . . . .
Indeterminateproducts
ExampleFind
limx→0+
√x ln x
SolutionJury-rigtheexpressiontomakeanindeterminatequotient. ThenapplyL’Hôpital’sRule:
limx→0+
√x ln x = lim
x→0+
ln x1/
√x
H= lim
x→0+
x−1
−12x
−3/2
= limx→0+
−2√x = 0
. . . . . .
Indeterminateproducts
ExampleFind
limx→0+
√x ln x
SolutionJury-rigtheexpressiontomakeanindeterminatequotient. ThenapplyL’Hôpital’sRule:
limx→0+
√x ln x = lim
x→0+
ln x1/
√x
H= lim
x→0+
x−1
−12x
−3/2
= limx→0+
−2√x = 0
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)
Thislimitisoftheform ∞−∞, whichisindeterminate.
SolutionAgain, rigittomakeanindeterminatequotient.
limx→0+
sin(2x) − x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
H= lim
x→0+
4x cos(2x)4 cos(2x) − 4x sin(2x)
= 0
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)Thislimitisoftheform ∞−∞, whichisindeterminate.
SolutionAgain, rigittomakeanindeterminatequotient.
limx→0+
sin(2x) − x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
H= lim
x→0+
4x cos(2x)4 cos(2x) − 4x sin(2x)
= 0
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)Thislimitisoftheform ∞−∞, whichisindeterminate.
SolutionAgain, rigittomakeanindeterminatequotient.
limx→0+
sin(2x) − x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
H= lim
x→0+
4x cos(2x)4 cos(2x) − 4x sin(2x)
= 0
. . . . . .
Indeterminatepowers
Example
limx→0+
(1− 2x)1/x
Takethelogarithm:
ln(limx→0
(1− 2x)1/x)
= limx→0
ln((1− 2x)1/x
)= lim
x→0
1xln(1− 2x)
Thislimitisoftheform00, sowecanuseL’Hôpital:
limx→0
1xln(1− 2x) H
= limx→0
−21−2x
1= −2
Thisisnottheanswer, it’sthelogoftheanswer! Sotheanswerwewantis e−2.
. . . . . .
Indeterminatepowers
Example
limx→0+
(1− 2x)1/x
Takethelogarithm:
ln(limx→0
(1− 2x)1/x)
= limx→0
ln((1− 2x)1/x
)= lim
x→0
1xln(1− 2x)
Thislimitisoftheform00, sowecanuseL’Hôpital:
limx→0
1xln(1− 2x) H
= limx→0
−21−2x
1= −2
Thisisnottheanswer, it’sthelogoftheanswer! Sotheanswerwewantis e−2.
. . . . . .
Example
limx→0
(3x)4x
Solution
ln limx→0+
(3x)4x = limx→0+
ln(3x)4x = limx→0+
4x ln(3x)
= limx→0+
ln(3x)1/4x
H= lim
x→0+
3/3x−1/4x2
= limx→0+
(−4x) = 0
Sotheansweris e0 = 1.
. . . . . .
Example
limx→0
(3x)4x
Solution
ln limx→0+
(3x)4x = limx→0+
ln(3x)4x = limx→0+
4x ln(3x)
= limx→0+
ln(3x)1/4x
H= lim
x→0+
3/3x−1/4x2
= limx→0+
(−4x) = 0
Sotheansweris e0 = 1.
. . . . . .
SummaryForm Method
00 L’Hôpital’sruledirectly
∞∞ L’Hôpital’sruledirectly
0 · ∞ jiggletomake 00 or ∞
∞ .
∞−∞ factortomakeanindeterminateproduct
00 take ln tomakeanindeterminateproduct
∞0 ditto
1∞ ditto
. . . . . .
Outline
Lasttime
IndeterminateForms
L’Hôpital’sRuleApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
TheCauchyMeanValueTheorem(Bonus)
Nexttime
. . . . . .
TheCauchyMeanValueTheorem(Bonus)
ApplytheMVT tothefunction
h(x) = (f(b) − f(a))g(x) − (g(b) − g(a))f(x).
Wehave h(a) = h(b). Sothereexistsa c in (a,b) suchthath′(c) = 0. Thus
(f(b) − f(a))g′(c) = (g(b) − g(a))f′(c)
ThisishowL’Hôpital’sRuleisproved.
. . . . . .
Outline
Lasttime
IndeterminateForms
L’Hôpital’sRuleApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
TheCauchyMeanValueTheorem(Bonus)
Nexttime
. . . . . .
NexttimeApplicationstoBusinessandEconomics
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