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Textbook by: Lawrence Baggett. E-mail the author

# L'Hopital's Rule

Module by: Lawrence Baggett. E-mail the author

Summary: The caught mean value theorem and L'Hopital's rule are included in this module, accompanied by practice exercises.

Many limits of certain combinations of functions are difficult to evaluate because they lead to what's known as “indeterminate forms.” These are expressions of the form 0/0,0/0,/,/,00,00,-,-,1,1, and the like. They are precisely combinations of functions that are not covered by our limit theorems. See (Reference). The very definition of the derivative itself is such a case: limh0(f(c+h)-f(c))=0,limh0(f(c+h)-f(c))=0,limh0h=0,limh0h=0, and we are interested in the limit of the quotient of these two functions, which would lead us to the indeterminate form 0/0.0/0. The definition of the number ee is another example: lim(1+1/n)=1,lim(1+1/n)=1,limn=,limn=, and we are interested in the limit of (1+1/n)n,(1+1/n)n, which leads to the indeterminate form 1.1. L'Hopital's Rule, Theorem 2 below, is our strongest tool for handling such indeterminate forms.

To begin with, here is a useful generalization of the Mean Value Theorem.

## Theorem 1: Cauchy Mean Value Theorem

Let ff and gg be continuous real-valued functions on a closed interval [a,b],[a,b], suppose g(a)g(b),g(a)g(b), and assume that both ff and gg are differentiable on the open interval (a,b).(a,b). Then there exists a point c(a,b)c(a,b) such that

f ( b ) - f ( a ) g ( b ) - g ( a ) = f ' ( c ) g ' ( c ) . f ( b ) - f ( a ) g ( b ) - g ( a ) = f ' ( c ) g ' ( c ) .
(1)

## Exercise 1

Prove the preceding theorem.

HINT: Define an auxiliary function hh as was done in the proof of the original Mean Value Theorem.

The following theorem and exercise comprise what is called L'Hopital's Rule.

## Theorem 2

Suppose ff and gg are differentiable real-valued functions on the bounded open interval (a,b)(a,b) and assume that

lim x a + 0 f ' ( x ) g ' ( x ) = L , lim x a + 0 f ' ( x ) g ' ( x ) = L ,
(2)

where LL is a real number. (Implicit in this hypothesis is that g'(x)0g'(x)0 for xx in some interval (a,a+α).(a,a+α).) Suppose further that either

lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = 0 lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = 0
(3)

or

lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = . lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = .
(4)

then

lim x a + 0 f ( x ) g ( x ) = L . lim x a + 0 f ( x ) g ( x ) = L .
(5)

### Proof

Suppose first that

lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = 0 . lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = 0 .
(6)

Observe first that, because g'(x)0g'(x)0 for all xx in some interval (a,a+α),(a,a+α),g'(x)g'(x) is either always positive or always negative on that interval. (This follows from part (d) of (Reference).) Therefore the function gg must be strictly monotonic on the interval (a,a+α).(a,a+α). Hence, since limxa+0g(x)=0,limxa+0g(x)=0, we must have that g(x)0g(x)0 on the interval (a,a+α).(a,a+α).

Now, given an ϵ>0,ϵ>0, choose a positive δ<αδ<α such that if a<c<a+δa<c<a+δ then |f'(c)g'(c)-L|<ϵ.|f'(c)g'(c)-L|<ϵ. Then, for every natural number nn for which 1/n<δ,1/n<δ, and every a<x<a+δ,a<x<a+δ, we have by the Cauchy Mean Value Theorem that there exists a point cc between a+1/na+1/n and xx such that

| f ( x ) - f ( a + 1 / n ) g ( x ) - g ( a + 1 / n ) - L | = | f ' ( c ) g ' ( c ) - L | < ϵ . | f ( x ) - f ( a + 1 / n ) g ( x ) - g ( a + 1 / n ) - L | = | f ' ( c ) g ' ( c ) - L | < ϵ .
(7)

Therefore, taking the limit as nn approaches ,, we obtain

| f ( x ) g ( x ) - L | = lim n | f ( x ) - f ( a + 1 / n ) g ( x ) - g ( a + 1 / n ) - L | ϵ | f ( x ) g ( x ) - L | = lim n | f ( x ) - f ( a + 1 / n ) g ( x ) - g ( a + 1 / n ) - L | ϵ
(8)

for all xx for which a<x<a+δ.a<x<a+δ. This proves the theorem in this first case.

Next, suppose that

lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = . lim x a + 0 f ( x ) = lim x a + 0 g ( x ) = .
(9)

This part of the theorem is a bit more complicated to prove. First, choose a positive αα so that f(x)f(x) and g(x)g(x) are both positive on the interval (a,a+α).(a,a+α). This is possible because both functions are tending to infinity as xx approaches a.a. Now, given an ϵ>0,ϵ>0, choose a positive number β<αβ<α such that

| f ' ( c ) g ' ( c ) - L | < ϵ 2 | f ' ( c ) g ' ( c ) - L | < ϵ 2
(10)

for all a<c<a+β.a<c<a+β. We express this absolute value inequality as the following pair of ordinary inequalities:

L - ϵ 2 < f ' ( c ) g ' ( c ) < L + ϵ 2 . L - ϵ 2 < f ' ( c ) g ' ( c ) < L + ϵ 2 .
(11)

Set y=a+β.y=a+β. Using the Cauchy Mean Value Theorem, and the preceding inequalities, we have that for all a<x<ya<x<y

L - ϵ 2 < f ( x ) - f ( y ) g ( x ) - g ( y ) < L + ϵ 2 , L - ϵ 2 < f ( x ) - f ( y ) g ( x ) - g ( y ) < L + ϵ 2 ,
(12)

implying that

( L - ϵ 2 ) ( g ( x ) - g ( y ) ) + f ( y ) < f ( x ) < ( L + ϵ 2 ) ( g ( x ) - g ( y ) ) + f ( y ) . ( L - ϵ 2 ) ( g ( x ) - g ( y ) ) + f ( y ) < f ( x ) < ( L + ϵ 2 ) ( g ( x ) - g ( y ) ) + f ( y ) .
(13)

Dividing through by g(x)g(x) and simplifying we obtain

L - ϵ 2 - ( L - ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) < f ( x ) g ( x ) < L + ϵ 2 - ( L + ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) . L - ϵ 2 - ( L - ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) < f ( x ) g ( x ) < L + ϵ 2 - ( L + ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) .
(14)

Finally, using the hypothesis that limxa+0g(x)=,limxa+0g(x)=, and the fact that L,ϵ,g(y),L,ϵ,g(y), and f(y)f(y) are all constants, choose a δ>0,δ>0, with δ<β,δ<β, such that if a<x<a+δ,a<x<a+δ, then

| - ( L - ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) | < ϵ 2 | - ( L - ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) | < ϵ 2
(15)

and

| - ( L + ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) | < ϵ 2 . | - ( L + ϵ 2 ) g ( y ) g ( x ) + f ( y ) g ( x ) | < ϵ 2 .
(16)

Then, for all a<x<a+δ,a<x<a+δ, we would have

L - ϵ < f ( x ) g ( x ) < L + ϵ , L - ϵ < f ( x ) g ( x ) < L + ϵ ,
(17)

implying that

| f ( x ) g ( x ) - L | < ϵ , | f ( x ) g ( x ) - L | < ϵ ,
(18)

and the theorem is proved.

## Exercise 2

1. Show that the conclusions of the preceding theorem also hold if we assume that
limxa+0f'(x)g'(x)=.limxa+0f'(x)g'(x)=.
(19)
HINT: Replace ϵϵ by a large real number BB and show that f(x)/g(x)>Bf(x)/g(x)>B if 0<x-a<δ.0<x-a<δ.
2. Show that the preceding theorem, as well as part (a) of this exercise, also holds if we replace the (finite) endpoint aa by -.-. HINT: Replace the δδ's by negative numbers B.B.
3. Show that the preceding theorem, as well as parts a and b of this exercise, hold if the limit as xx approaches aa from the right is replaced by the limit as xx approaches bb from the left. HINT: Replace f(x)f(x) by f(-x)f(-x) and g(x)g(x) by g(-x).g(-x).
4. Give an example to show that the converse of L'Hopital's Rule need not hold; i.e., find functions ff and gg for which limxa+0f(x)=limxa+0g(x)=0,limxa+0f(x)=limxa+0g(x)=0,
limxa+0f(x)g(x)exists,butlimxa+0f'(x)g'(x)doesnotexist.limxa+0f(x)g(x)exists,butlimxa+0f'(x)g'(x)doesnotexist.
(20)
5. Deduce from the proof given above that if limxa+0f'(x)/g'(x)=Llimxa+0f'(x)/g'(x)=L and limxa+0g(x)=,limxa+0g(x)=, then limxa+0f(x)/g(x)=Llimxa+0f(x)/g(x)=L independent of the behavior of f.f.
6. Evaluate limxx1/x,limxx1/x, and limx0(1-x)1/x.limx0(1-x)1/x. HINT: Take logarithms.

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