lesson 18: indeterminate forms and l'hôpital's rule
DESCRIPTION
L'Hôpital's Rule is not a magic bullet (or a sledgehammer) but it does allow us to find limits of indeterminate forms such as 0/0 and ∞/∞. With some algebra we can use it to resolve other indeterminate forms such as ∞-∞ and 0^0.TRANSCRIPT
. . . . . .
Section3.7IndeterminateFormsandL’Hôpital’s
Rule
V63.0121.034, CalculusI
November2, 2009
Announcements
I NoofficehourstodayI Springforward, fallback, yaddayaddayadda
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Experimentswithfunnylimits
I limx→0+
sin2 xx
= 0
I limx→0
xsin2 x
doesnotexist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
Alloftheseareoftheform00, andsincewecangetdifferent
answersindifferentcases, wesaythisformis indeterminate.
. . . . . .
Outline
IndeterminateForms
L’Hôpital’sRuleRelativeRatesofGrowthApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
I Limitofasumisthesumofthelimits
I LimitofadifferenceisthedifferenceofthelimitsI LimitofaproductistheproductofthelimitsI Limitofaquotientisthequotientofthelimits... whoops!Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
I LimitofasumisthesumofthelimitsI Limitofadifferenceisthedifferenceofthelimits
I LimitofaproductistheproductofthelimitsI Limitofaquotientisthequotientofthelimits... whoops!Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
I LimitofasumisthesumofthelimitsI LimitofadifferenceisthedifferenceofthelimitsI Limitofaproductistheproductofthelimits
I Limitofaquotientisthequotientofthelimits... whoops!Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
Recall
RecallthelimitlawsfromChapter2.
I LimitofasumisthesumofthelimitsI LimitofadifferenceisthedifferenceofthelimitsI LimitofaproductistheproductofthelimitsI Limitofaquotientisthequotientofthelimits... whoops!Thisistrueaslongasyoudon’ttrytodividebyzero.
. . . . . .
I Weknowdividingbyzeroisbad.I Mostofthetime, ifanexpression’snumeratorapproachesafinitenumberanddenominatorapproacheszero, thequotientapproachessomekindofinfinity. Forexample:
limx→0+
1x
= +∞ limx→0−
cos xx3
= −∞
I Anexceptionwouldbesomethinglike
limx→∞
11x sin x
= limx→∞
x csc x.
whichdoesn’texist.I Evenlesspredictable: numeratoranddenominatorbothgotozero.
. . . . . .
I Weknowdividingbyzeroisbad.I Mostofthetime, ifanexpression’snumeratorapproachesafinitenumberanddenominatorapproacheszero, thequotientapproachessomekindofinfinity. Forexample:
limx→0+
1x
= +∞ limx→0−
cos xx3
= −∞
I Anexceptionwouldbesomethinglike
limx→∞
11x sin x
= limx→∞
x csc x.
whichdoesn’texist.
I Evenlesspredictable: numeratoranddenominatorbothgotozero.
. . . . . .
I Weknowdividingbyzeroisbad.I Mostofthetime, ifanexpression’snumeratorapproachesafinitenumberanddenominatorapproacheszero, thequotientapproachessomekindofinfinity. Forexample:
limx→0+
1x
= +∞ limx→0−
cos xx3
= −∞
I Anexceptionwouldbesomethinglike
limx→∞
11x sin x
= limx→∞
x csc x.
whichdoesn’texist.I Evenlesspredictable: numeratoranddenominatorbothgotozero.
. . . . . .
LanguageNoteItdependsonwhatthemeaningoftheword“is”is
I Becarefulwiththelanguagehere. Wearenot sayingthatthelimit
ineachcase“is”00, and
thereforenonexistentbecausethisexpressionisundefined.
I Thelimit isoftheform00, whichmeanswe
cannotevaluateitwithourlimitlaws.
. . . . . .
IndeterminateformsarelikeTugOfWar
Whichsidewinsdependsonwhichsideisstronger.
. . . . . .
Outline
IndeterminateForms
L’Hôpital’sRuleRelativeRatesofGrowthApplicationtoIndeterminateProductsApplicationtoIndeterminateDifferencesApplicationtoIndeterminatePowersSummary
. . . . . .
TheLinearCase
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
. . . . . .
TheLinearCase
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
. . . . . .
TheLinearCase, Illustrated
. .x
.y
.y = f(x)
.y = g(x)
..a
..x
.f(x).g(x)
f(x)g(x)
=f(x) − f(a)g(x) − g(a)
=(f(x) − f(a))/(x− a)(g(x) − g(a))/(x− a)
=m1
m2
. . . . . .
Whatthen?
I Butwhatifthefunctionsaren’tlinear?
I Canweapproximateafunctionnearapointwithalinearfunction?
I Whatwouldbetheslopeofthatlinearfunction?
. . . . . .
Whatthen?
I Butwhatifthefunctionsaren’tlinear?I Canweapproximateafunctionnearapointwithalinearfunction?
I Whatwouldbetheslopeofthatlinearfunction?
. . . . . .
Whatthen?
I Butwhatifthefunctionsaren’tlinear?I Canweapproximateafunctionnearapointwithalinearfunction?
I Whatwouldbetheslopeofthatlinearfunction?
. . . . . .
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 −∞.
. . . . . .
MeettheMathematician: L’Hôpital
I wantedtobeamilitaryman, butpooreyesightforcedhimintomath
I didsomemathonhisown(solvedthe“brachistocroneproblem”)
I paidastipendtoJohannBernoulli, whoprovedthistheoremandnameditafterhim! GuillaumeFrançoisAntoine,
MarquisdeL’Hôpital(1661–1704)
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x.
.numerator → 0
sin x2.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x.
.numerator → 0
sin x2.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)
H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)
H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x.
.numerator → 0
(cos x2) (�2x.
.denominator → 0
)
H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x.
.numerator → 0
(cos x2) (�2x.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x.
.numerator → 1
cos x2 − 2x2 sin(x2).
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
Revisitingthepreviousexamples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
.
.sin x → 0
cos x1
= 0
Example
limx→0
sin2 x
.
.numerator → 0
sin x2
.
.denominator → 0
H= lim
x→0
�2 sin x cos x
.
.numerator → 0
(cos x2) (�2x
.
.denominator → 0
)H= lim
x→0
cos2 x− sin2 x
.
.numerator → 1
cos x2 − 2x2 sin(x2)
.
.denominator → 1
= 1
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
. . . . . .
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!= ∞.
Forexample, if r = 3, threeinvocationsofL’Hôpital’sRulegiveus
limx→∞
ex
x3H= lim
x→∞
ex
3 · x2H= lim
x→∞
ex
2 · 3xH= lim
x→∞
ex
1 · 2 · 3= ∞
. . . . . .
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!= ∞.
Forexample, if r = 3, threeinvocationsofL’Hôpital’sRulegiveus
limx→∞
ex
x3H= lim
x→∞
ex
3 · x2H= lim
x→∞
ex
2 · 3xH= lim
x→∞
ex
1 · 2 · 3= ∞
. . . . . .
If r isnotaninteger, let m bethesmallestintegergreaterthan r.Thenif x > 1, xr < xm, so
ex
xr>
ex
xm.
Theright-handsidetendsto ∞ bythepreviousstep.
Forexample, if r = 1/2, r < 1 sofor x > 1
ex
x1/2>
ex
x
whichgetsarbitrarilylarge.
. . . . . .
If r isnotaninteger, let m bethesmallestintegergreaterthan r.Thenif x > 1, xr < xm, so
ex
xr>
ex
xm.
Theright-handsidetendsto ∞ bythepreviousstep. Forexample, if r = 1/2, r < 1 sofor x > 1
ex
x1/2>
ex
x
whichgetsarbitrarilylarge.
. . . . . .
TheoremLet r beanypositivenumber. Then
limx→∞
ln xxr
= 0.
Proof.OneapplicationofL’Hôpital’sRuleheresuffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
. . . . . .
TheoremLet r beanypositivenumber. Then
limx→∞
ln xxr
= 0.
Proof.OneapplicationofL’Hôpital’sRuleheresuffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
. . . . . .
Indeterminateproducts
ExampleFind
limx→0+
√x ln x
Thislimitisoftheform 0 · (−∞).
SolutionJury-rigtheexpressiontomakeanindeterminatequotient. ThenapplyL’Hôpital’sRule:
limx→0+
√x ln x
= limx→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
Thislimitisoftheform 0 · (−∞).
SolutionJury-rigtheexpressiontomakeanindeterminatequotient. ThenapplyL’Hôpital’sRule:
limx→0+
√x ln x
= limx→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
Thislimitisoftheform 0 · (−∞).
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
Thislimitisoftheform 0 · (−∞).
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
Thislimitisoftheform 0 · (−∞).
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
Thislimitisoftheform 0 · (−∞).
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 ∞−∞.
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)
= ∞
Thelimitis +∞ becuasethenumeratortendsto 1 whilethedenominatortendstozerobutremainspositive.
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)Thislimitisoftheform ∞−∞.
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)
= ∞
Thelimitis +∞ becuasethenumeratortendsto 1 whilethedenominatortendstozerobutremainspositive.
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)Thislimitisoftheform ∞−∞.
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)
= ∞
Thelimitis +∞ becuasethenumeratortendsto 1 whilethedenominatortendstozerobutremainspositive.
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)Thislimitisoftheform ∞−∞.
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)
= ∞
Thelimitis +∞ becuasethenumeratortendsto 1 whilethedenominatortendstozerobutremainspositive.
. . . . . .
Indeterminatedifferences
Example
limx→0
(1x− cot 2x
)Thislimitisoftheform ∞−∞.
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)
= ∞
Thelimitis +∞ becuasethenumeratortendsto 1 whilethedenominatortendstozerobutremainspositive.
. . . . . .
Checkingyourwork
.
.limx→0
tan 2x2x
= 1, so for small x,
tan 2x ≈ 2x. So cot 2x ≈ 12x
and
1x− cot 2x ≈ 1
x− 1
2x=
12x
→ ∞
as x → 0+.
. . . . . .
Indeterminatepowers
ExampleFind lim
x→0+(1− 2x)1/x
Takethelogarithm:
ln(limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+
ln(1− 2x)x
Thislimitisoftheform00, sowecanuseL’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x
1= −2
Thisisnottheanswer, it’sthelogoftheanswer! Sotheanswerwewantis e−2.
. . . . . .
Indeterminatepowers
ExampleFind lim
x→0+(1− 2x)1/x
Takethelogarithm:
ln(limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+
ln(1− 2x)x
Thislimitisoftheform00, sowecanuseL’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x
1= −2
Thisisnottheanswer, it’sthelogoftheanswer! Sotheanswerwewantis e−2.
. . . . . .
Indeterminatepowers
ExampleFind lim
x→0+(1− 2x)1/x
Takethelogarithm:
ln(limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+
ln(1− 2x)x
Thislimitisoftheform00, sowecanuseL’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x
1= −2
Thisisnottheanswer, it’sthelogoftheanswer! Sotheanswerwewantis e−2.
. . . . . .
Indeterminatepowers
ExampleFind lim
x→0+(1− 2x)1/x
Takethelogarithm:
ln(limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+
ln(1− 2x)x
Thislimitisoftheform00, sowecanuseL’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→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
. . . . . .
Finalthoughts
I L’Hôpital’sRuleonlyworksonindeterminatequotientsI Luckily, mostindeterminatelimitscanbetransformedintoindeterminatequotients
I L’Hôpital’sRulegiveswronganswersfornon-indeterminatelimits!