lesson 25: indeterminate forms and l'hôpital's rule
DESCRIPTION
Recognizing indeterminate forms and resolving them with L'Hôpital's RuleTRANSCRIPT
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Section 4.5Indeterminate Forms and L’Hopital’s Rule
Math 1a
November 26, 2007
Announcements
I Special review session on optimization problems:Tues 11/27 (tomorrow) 7:00–9:00 (SC 507)
I my next office hours: today 1–2, tomorrow 3–4 (SC 323)
I MT II Review session: Sunday, 11/2, 7:30–9:00 (SC Hall D)
I Midterm II: Tues 12/4 7:00-9:00pm (SC Hall B)
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Outline
Indeterminate Forms
L’Hopital’s RuleApplication to Indeterminate ProductsApplication to Indeterminate DifferencesApplication to Indeterminate PowersSummary
The Cauchy Mean Value Theorem (Bonus)
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Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limits
I Limit of a difference is the difference of the limits
I Limit of a product is the product of the limits
I Limit of a quotient is the quotient of the limits ... whoops!This is true as long as you don’t try to divide by zero.
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Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limits
I Limit of a difference is the difference of the limits
I Limit of a product is the product of the limits
I Limit of a quotient is the quotient of the limits ... whoops!This is true as long as you don’t try to divide by zero.
![Page 5: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/5.jpg)
Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limits
I Limit of a difference is the difference of the limits
I Limit of a product is the product of the limits
I Limit of a quotient is the quotient of the limits ... whoops!This is true as long as you don’t try to divide by zero.
![Page 6: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/6.jpg)
Recall
Recall the limit laws from Chapter 2.
I Limit of a sum is the sum of the limits
I Limit of a difference is the difference of the limits
I Limit of a product is the product of the limits
I Limit of a quotient is the quotient of the limits ... whoops!This is true as long as you don’t try to divide by zero.
![Page 7: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/7.jpg)
We know dividing by zero is bad. Most of the time, if you have anumerator which approaches a finite number and a denominatorwhich approaches zero, the quotient approaches some kind ofinfinity. An exception would be something like
limx→∞
11x sin x
= limx→∞
x sec x .
which doesn’t exist.
Even worse is the situation where the numerator and denominatorboth go to zero.
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We know dividing by zero is bad. Most of the time, if you have anumerator which approaches a finite number and a denominatorwhich approaches zero, the quotient approaches some kind ofinfinity. An exception would be something like
limx→∞
11x sin x
= limx→∞
x sec x .
which doesn’t exist.Even worse is the situation where the numerator and denominatorboth go to zero.
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Experiments
I limx→0+
sin2 x
x
= 0
I limx→0
x
sin2 x
does not exist
I limx→0
sin2 x
sin x2
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 x
does not exist
I limx→0
sin2 x
sin x2
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 x
does not exist
I limx→0
sin2 x
sin x2
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin x2
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin x2
= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin x2= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin x2= 1
I limx→0
sin 3x
sin x
= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin x2= 1
I limx→0
sin 3x
sin x= 3
All of these are of the form0
0, and since we can get different
answers in different cases, we say this form is indeterminate.
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Experiments
I limx→0+
sin2 x
x= 0
I limx→0
x
sin2 xdoes not exist
I limx→0
sin2 x
sin x2= 1
I limx→0
sin 3x
sin x= 3
All of these are of the form0
0, 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. We are not saying that the
limit in each case “is”0
0, and therefore nonexistent because
this expression is undefined.
I The limit is of the form0
0, which means we cannot evaluate it
with our limit laws.
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Indeterminate forms are like Tug Of War
Which side wins depends on which side is stronger.
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Outline
Indeterminate Forms
L’Hopital’s RuleApplication to Indeterminate ProductsApplication to Indeterminate DifferencesApplication to Indeterminate PowersSummary
The Cauchy Mean Value Theorem (Bonus)
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QuestionIf f and g are lines and f (a) = g(a) = 0, what is
limx→a
f (x)
g(x)?
SolutionThe functions f and g can be written in the form
f (x) = m1(x − a)
g(x) = m2(x − a)
Sof (x)
g(x)=
m1
m2=
f ′(x)
g ′(x).
But what if the functions aren’t linear? If only there were a way todeal with functions which were only approximately linear!
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QuestionIf f and g are lines and f (a) = g(a) = 0, what is
limx→a
f (x)
g(x)?
SolutionThe functions f and g can be written in the form
f (x) = m1(x − a)
g(x) = m2(x − a)
Sof (x)
g(x)=
m1
m2=
f ′(x)
g ′(x).
But what if the functions aren’t linear? If only there were a way todeal with functions which were only approximately linear!
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QuestionIf f and g are lines and f (a) = g(a) = 0, what is
limx→a
f (x)
g(x)?
SolutionThe functions f and g can be written in the form
f (x) = m1(x − a)
g(x) = m2(x − a)
Sof (x)
g(x)=
m1
m2=
f ′(x)
g ′(x).
But what if the functions aren’t linear? If only there were a way todeal with functions which were only approximately linear!
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Theorem (L’Hopital’s Rule)
Suppose f and g are differentiable functions and g ′(x) 6= 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) = ±∞
Then
limx→a
f (x)
g(x)= lim
x→a
f ′(x)
g ′(x),
if the limit on the right-hand side is finite, ∞, or −∞.
L’Hopital’s rule also applies for limits of the form∞∞
.
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Theorem (L’Hopital’s Rule)
Suppose f and g are differentiable functions and g ′(x) 6= 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) = ±∞
Then
limx→a
f (x)
g(x)= lim
x→a
f ′(x)
g ′(x),
if the limit on the right-hand side is finite, ∞, or −∞.
L’Hopital’s rule also applies for limits of the form∞∞
.
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Meet the Mathematician
I wanted to be a militaryman, but poor eyesightforced him into math
I did some math on hisown (solved the“brachistocroneproblem”)
I paid a stipend to JohannBernoulli, who provedthis theorem and namedit after him! Guillaume Franois Antoine,
Marquis de L’Hopital(1661–1704)
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How does this affect our examples above?
Example
limx→0
sin2 x
x
H= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
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How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
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How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
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How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)
H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
![Page 31: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/31.jpg)
How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)
= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
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How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
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How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin x
H= lim
x→0
3 cos 3x
cos x= 3.
![Page 34: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/34.jpg)
How does this affect our examples above?
Example
limx→0
sin2 x
xH= lim
x→0
2 sin x cos x
1= 0.
Example
limx→0
sin2 x
sin x2
H= lim
x→0
�2 sin x cos x
(cos x2) (�2x)H= lim
x→0
cos2 x − sin2 x
cos x2 − x2 sin(x2)= 1
Example
limx→0
sin 3x
sin xH= lim
x→0
3 cos 3x
cos x= 3.
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Sketch of Proof of L’Hopital’s Rule
Let x be a number close to a. We know thatf (x)− f (a)
x − a= f ′(c),
for some c ∈ (a, x); alsog(x)− g(a)
x − a= g ′(d), for some
d ∈ (a, x). This means
f (x)
g(x)≈ f ′(c)
g ′(d).
The miracle of the MVT is that a tweaking of it allows us toassume c = d , so that
f (x)
g(x)=
f ′(c)
g ′(c).
The number c depends on x and since it is between a and x , wemust have
limx→a
c(x) = a.
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Beware of Red Herrings
Example
Findlimx→0
x
cos x
SolutionThe limit of the denominator is 1, not 0, so L’Hopital’s rule doesnot apply. The limit is 0.
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Beware of Red Herrings
Example
Findlimx→0
x
cos x
SolutionThe limit of the denominator is 1, not 0, so L’Hopital’s rule doesnot apply. The limit is 0.
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TheoremLet r be any positive number. Then
limx→∞
ex
x r=∞.
Proof.If r is a positive integer, then apply L’Hopital’s rule r times to thefraction. You get
limx→∞
ex
x r
H= . . .
H= lim
x→∞
ex
r !=∞.
If r is not an integer, let n = [[x ]] and m = n + 1. Then if x > 1,xn < x r < xm, so
ex
xn>
ex
x r>
ex
xm.
Now apply the Squeeze Theorem.
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TheoremLet r be any positive number. Then
limx→∞
ex
x r=∞.
Proof.If r is a positive integer, then apply L’Hopital’s rule r times to thefraction. You get
limx→∞
ex
x r
H= . . .
H= lim
x→∞
ex
r !=∞.
If r is not an integer, let n = [[x ]] and m = n + 1. Then if x > 1,xn < x r < xm, so
ex
xn>
ex
x r>
ex
xm.
Now apply the Squeeze Theorem.
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TheoremLet r be any positive number. Then
limx→∞
ex
x r=∞.
Proof.If r is a positive integer, then apply L’Hopital’s rule r times to thefraction. You get
limx→∞
ex
x r
H= . . .
H= lim
x→∞
ex
r !=∞.
If r is not an integer, let n = [[x ]] and m = n + 1. Then if x > 1,xn < x r < xm, so
ex
xn>
ex
x r>
ex
xm.
Now apply the Squeeze Theorem.
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Indeterminate products
Example
Findlim
x→0+
√x ln x
SolutionJury-rig the expression to make an indeterminate quotient. Thenapply L’Hopital’s Rule:
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
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Indeterminate products
Example
Findlim
x→0+
√x ln x
SolutionJury-rig the expression to make an indeterminate quotient. Thenapply L’Hopital’s Rule:
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
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Indeterminate differences
Example
limx→0
(1
x− cot 2x
)
This limit is of the form ∞−∞, which is indeterminate.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
1− x cot 2x
xH= lim
x→0+
2x csc2(2x)− cot(2x)
1= lim
x→0+
2x − cos 2x
sin2 x sin 2xH= lim
x→0+
2 + 2 sin 2x
2 cos 2x sin2 x + 2 cos x sin x sin 2x
=∞
![Page 44: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/44.jpg)
Indeterminate differences
Example
limx→0
(1
x− cot 2x
)This limit is of the form ∞−∞, which is indeterminate.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
1− x cot 2x
xH= lim
x→0+
2x csc2(2x)− cot(2x)
1= lim
x→0+
2x − cos 2x
sin2 x sin 2xH= lim
x→0+
2 + 2 sin 2x
2 cos 2x sin2 x + 2 cos x sin x sin 2x
=∞
![Page 45: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/45.jpg)
Indeterminate differences
Example
limx→0
(1
x− cot 2x
)This limit is of the form ∞−∞, which is indeterminate.
SolutionAgain, rig it to make an indeterminate quotient.
limx→0+
1− x cot 2x
xH= lim
x→0+
2x csc2(2x)− cot(2x)
1= lim
x→0+
2x − cos 2x
sin2 x sin 2xH= lim
x→0+
2 + 2 sin 2x
2 cos 2x sin2 x + 2 cos x sin x sin 2x
=∞
![Page 46: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/46.jpg)
Indeterminate powers
Example
limx→0+
(1− 2x)1/x
Take the logarithm:
ln(
limx→0
(1− 2x)1/x)
= limx→0
ln(
(1− 2x)1/x)
= limx→0
1
xln(1− 2x)
This limit is of the form0
0, so we can use L’Hopital:
limx→0
1
xln(1− 2x)
H= lim
x→0
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answerwe want is e−2.
![Page 47: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/47.jpg)
Indeterminate powers
Example
limx→0+
(1− 2x)1/x
Take the logarithm:
ln(
limx→0
(1− 2x)1/x)
= limx→0
ln(
(1− 2x)1/x)
= limx→0
1
xln(1− 2x)
This limit is of the form0
0, so we can use L’Hopital:
limx→0
1
xln(1− 2x)
H= lim
x→0
−21−2x
1= −2
This is not the answer, it’s the log of the answer! So the answerwe want is e−2.
![Page 48: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/48.jpg)
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
So the answer is e0 = 1.
![Page 49: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/49.jpg)
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
So the answer is e0 = 1.
![Page 50: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/50.jpg)
Summary
Form Method
00 L’Hopital’s rule directly
∞∞ L’Hopital’s rule directly
0 · ∞ jiggle to make 00 or ∞∞ .
∞−∞ factor to make an indeterminate product
00 take ln to make an indeterminate product
∞0 ditto
1∞ ditto
![Page 51: Lesson 25: Indeterminate Forms and L'Hôpital's Rule](https://reader033.vdocuments.site/reader033/viewer/2022052508/5595aed81a28abe83d8b45d5/html5/thumbnails/51.jpg)
Outline
Indeterminate Forms
L’Hopital’s RuleApplication to Indeterminate ProductsApplication to Indeterminate DifferencesApplication to Indeterminate PowersSummary
The Cauchy Mean Value Theorem (Bonus)
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The Cauchy Mean Value Theorem (Bonus)
Apply the MVT to the function
h(x) = (f (b)− f (a))g(x)− (g(b)− g(a))f (x).
We have h(a) = h(b). So there exists a c in (a, b) such thath′(c) = 0. Thus
(f (b)− f (a))g ′(c) = (g(b)− g(a))f ′(c)
This is how L’Hopitalis proved.