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L'Hôpital's Rule allows us to resolve limits of indeterminate form: 0/0, ∞/∞, ∞-∞, 0^0, 1^∞, and ∞^0
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Sec on 3.7Indeterminate forms and lHôpital’s
Rule
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
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Announcements
I Midterm has beenreturned. Please see FAQon Blackboard (under”Exams and Quizzes”)
I Quiz 3 this week inrecita on on Sec on 2.6,2.8, 3.1, 3.2
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ObjectivesI Know when a limit is ofindeterminate form:
I indeterminate quo ents:0/0,∞/∞
I indeterminate products:0×∞
I indeterminatedifferences: ∞−∞
I indeterminate powers:00,∞0, and 1∞
I Resolve limits inindeterminate form
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Notes
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Notes
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Notes
. 1.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Recall
Recall the limit laws from Chapter 2.I Limit of a sum is the sum of the limitsI Limit of a difference is the difference of the limitsI Limit of a product is the product of the limitsI Limit of a quo ent is the quo ent of the limits ... whoops! Thisis true as long as you don’t try to divide by zero.
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More about dividing limits
I We know dividing by zero is bad.I Most of the me, if an expression’s numerator approaches afinite nonzero number and denominator approaches zero, thequo ent has an infinite. For example:
limx→0+
1x= +∞ lim
x→0−
cos xx3
= −∞
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Why 1/0 ̸= ∞Consider the func on f(x) =
11x sin x
.
..x
.
y
Then limx→∞
f(x) is of the form 1/0, but the limit does not exist and isnot infinite.Even less predictable: when numerator and denominator both go tozero.
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Notes
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Notes
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Notes
. 2.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
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All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
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Notes
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Notes
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Notes
. 3.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Experiments with funny limitsI lim
x→0
sin2 xx
= 0
I limx→0
xsin2 x
does not exist
I limx→0
sin2 xsin(x2)
= 1
I limx→0
sin 3xsin x
= 3
.
All of these are of the form00, and since we can get different
answers in different cases, we say this form is indeterminate.
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Language NoteIt depends on what the meaning of the word “is” is
I Be careful with the language here. Weare not saying that the limit in eachcase “is”
00, and therefore nonexistent
because this expression is undefined.
I The limit is of the form00, which means
we cannot evaluate it with our limitlaws.
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Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
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Notes
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Notes
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Notes
. 4.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
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The Linear CaseQues on
If f and g are lines and f(a) = g(a) = 0, what is limx→a
f(x)g(x)
?
Solu onThe func ons f and g can be wri en in the form
f(x) = m1(x− a) g(x) = m2(x− a)
Sof(x)g(x)
=m1
m2
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The Linear Case, 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
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Notes
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Notes
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Notes
. 5.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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What then?
I But what if the func ons aren’t linear?I Can we approximate a func on near a point with a linearfunc on?
I What would be the slope of that linear func on?
Thederiva ve!
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Theorem of the DayTheorem (L’Hopital’s Rule)
Suppose f and g are differen able func ons and g′(x) ̸= 0 near a(except possibly at a). Suppose that
limx→a
f(x) = 0 and limx→a
g(x) = 0
or limx→a
f(x) = ±∞ and limx→a
g(x) = ±∞
Thenlimx→a
f(x)g(x)
= limx→a
f′(x)g′(x)
,
if the limit on the right-hand side is finite,∞, or−∞.
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Meet the Mathematician
I wanted to be a military man, butpoor eyesight forced him intomath
I did some math on his own(solved the “brachistocroneproblem”)
I paid a s pend to JohannBernoulli, who proved thistheorem and named it a er him!
Guillaume François Antoine,Marquis de L’Hôpital(French, 1661–1704)
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Notes
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Notes
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Notes
. 6.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Revisiting the previous examples
Example
limx→0
sin2 xx
H= lim
x→0
2 sin x
..
sin x → 0
cos x1
= 0
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Revisiting the previous examples
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
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Revisiting the previous examples
Example
limx→0
sin 3xsin x
H= lim
x→0
3 cos 3xcos x
= 3.
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Notes
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Notes
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Notes
. 7.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Beware of Red Herrings
Example
Findlimx→0
xcos x
Solu on
The limit of the denominator is 1, not 0, so L’Hôpital’s rule does notapply. The limit is 0.
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Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
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Limits of Rational Functionsrevisited
Example
Find limx→∞
5x2 + 3x− 13x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 13x2 + 7x+ 27
H= lim
x→∞
10x+ 36x+ 7
H= lim
x→∞
106
=53
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Notes
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Notes
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Notes
. 8.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Limits of Rational Functionsrevisited II
Example
Find limx→∞
5x2 + 3x− 17x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
5x2 + 3x− 17x+ 27
H= lim
x→∞
10x+ 37
= ∞
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Limits of Rational Functionsrevisited III
Example
Find limx→∞
4x+ 73x2 + 7x+ 27
if it exists.
Solu onUsing L’Hôpital:
limx→∞
4x+ 73x2 + 7x+ 27
H= lim
x→∞
46x+ 7
= 0
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Limits of Rational FunctionsFactLet f(x) and g(x) be polynomials of degree p and q.
I If p > q, then limx→∞
f(x)g(x)
= ∞
I If p < q, then limx→∞
f(x)g(x)
= 0
I If p = q, then limx→∞
f(x)g(x)
is the ra o of the leading coefficients of
f and g.
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Notes
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Notes
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Notes
. 9.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Exponential vs. geometric growth
Example
Find limx→∞
ex
x2, if it exists.
Solu onWe have
limx→∞
ex
x2H= lim
x→∞
ex
2xH= lim
x→∞
ex
2= ∞.
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Exponential vs. geometric growthExample
What about limx→∞
ex
x3?
Answer
S ll∞. (Why?)
Solu on
limx→∞
ex
x3H= lim
x→∞
ex
3x2H= lim
x→∞
ex
6xH= lim
x→∞
ex
6= ∞.
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Exponential vs. fractional powersExample
Find limx→∞
ex√x, if it exists.
Solu on (without L’Hôpital)
We have for all x > 1, x1/2 < x1, so
ex
x1/2>
ex
x
The right hand side tends to∞, so the le -hand side must, too.
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Notes
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Notes
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Notes
. 10.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Exponential vs. fractional powersExample
Find limx→∞
ex√x, if it exists.
Solu on (with L’Hôpital)
limx→∞
ex√x= lim
x→∞
ex12x−1/2 = lim
x→∞2√xex = ∞
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Exponential vs. any powerTheorem
Let r be any posi ve number. Then limx→∞
ex
xr= ∞.
Proof.If r is a posi ve integer, then apply L’Hôpital’s rule r mes to the frac-on. You get
limx→∞
ex
xrH= . . .
H= lim
x→∞
ex
r!= ∞.
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Exponential vs. any powerTheorem
Let r be any posi ve number. Then limx→∞
ex
xr= ∞.
Proof.If r is not an integer, letm be the smallest integer greater than r. Thenif x > 1, xr < xm, so
ex
xr>
ex
xm. The right-hand side tends to∞ by the
previous step.
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Notes
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Notes
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Notes
. 11.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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Any exponential vs. any powerTheorem
Let a > 1 and r > 0. Then limx→∞
ax
xr= ∞.
Proof.If r is a posi ve integer, we have
limx→∞
ax
xrH= . . .
H= lim
x→∞
(ln a)rax
r!= ∞.
If r isn’t an integer, we can compare it as before.
So even limx→∞
(1.00000001)x
x100000000= ∞!
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Logarithmic versus power growthTheorem
Let r be any posi ve number. Then limx→∞
ln xxr
= 0.
Proof.One applica on of L’Hôpital’s Rule here suffices:
limx→∞
ln xxr
H= lim
x→∞
1/xrxr−1 = lim
x→∞
1rxr
= 0.
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Outline
L’Hôpital’s Rule
Rela ve Rates of Growth
Other Indeterminate LimitsIndeterminate ProductsIndeterminate DifferencesIndeterminate Powers
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Notes
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Notes
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Notes
. 12.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
.
.
Indeterminate productsExample
Find limx→0+
√x ln x
This limit is of the form 0 · (−∞).
Solu onJury-rig the expression to make an indeterminate quo ent. Thenapply L’Hôpital’s Rule:
limx→0+
√x ln x = lim
x→0+
ln x1/√x
H= lim
x→0+
x−1
−12x−3/2 = lim
x→0+−2
√x = 0
.
Indeterminate differences
Example
limx→0+
(1x− cot 2x
)This limit is of the form∞−∞.
.
Indeterminate DifferencesSolu on
Again, rig it to make an indeterminate quo ent.
limx→0+
(1x− cot 2x
)= lim
x→0+
sin(2x)− x cos(2x)x sin(2x)
H= lim
x→0+
cos(2x) + 2x sin(2x)2x cos(2x) + sin(2x)
= ∞
The limit is +∞ because the numerator tends to 1 while thedenominator tends to zero but remains posi ve.
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Notes
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Notes
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Notes
. 13.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
.
.
Indeterminate powersExample
Find limx→0+
(1− 2x)1/x
Solu onTake the logarithm:
ln(
limx→0+
(1− 2x)1/x)
= limx→0+
ln((1− 2x)1/x
)= lim
x→0+
ln(1− 2x)x
.
Indeterminate powersExample
Find limx→0+
(1− 2x)1/x
Solu on
This limit is of the form00, so we can use L’Hôpital:
limx→0+
ln(1− 2x)x
H= lim
x→0+
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answer wewant is e−2.
.
Another indeterminate power limitExample
Find limx→0
(3x)4x
Solu on
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
So the answer is e0 = 1.
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Notes
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Notes
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Notes
. 14.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .
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SummaryForm Method
00 L’Hôpital’s rule directly∞∞ L’Hôpital’s rule directly
0 · ∞ jiggle to make 00 or
∞∞ .
∞−∞ combine to make an indeterminate product or quo ent
00 take ln to make an indeterminate product
∞0 di o
1∞ di o
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Final Thoughts
I L’Hôpital’s Rule only works on indeterminate quo entsI Luckily, most indeterminate limits can be transformed intoindeterminate quo ents
I L’Hôpital’s Rule gives wrong answers for non-indeterminatelimits!
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Notes
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Notes
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Notes
. 15.
. Sec on 3.7: Indeterminate forms and lHôpital’s Rule. V63.0121.001: Calculus I .