5.17. L’HÔPITAL’S RULE 147

Now pick x,y such that c < x < y < b. By the Cauchy mean value theorem, there existst ∈ (x,y) such that

g′ (t)( f (x)− f (y)) = f ′ (t)(g(x)−g(y)) .

Since g′ (s) ̸= 0 for all s ∈ (a,b) it follows from the mean value theorem g(x)−g(y) ̸= 0.Therefore,

f ′ (t)g′ (t)

=f (x)− f (y)g(x)−g(y)

and so, since t > c, ∣∣∣∣ f ′ (t)g′ (t)

−L∣∣∣∣= ∣∣∣∣ f (x)− f (y)

g(x)−g(y)−L∣∣∣∣< ε

2.

Now taking limy→b−, ∣∣∣∣ f (x)g(x)

−L∣∣∣∣≤ ε

2< ε.

Since ε > 0 is arbitrary, this shows 5.10.The following corollary is proved in the same way.

Corollary 5.17.2 Let [a,b]⊆ [−∞,∞] and suppose f ,g are functions which satisfy,

limx→a+

f (x) = limx→a+

g(x) = 0, (5.11)

and f ′ and g′ exist on (a,b) with g′ (x) ̸= 0 on (a,b). Suppose also that

limx→a+

f ′ (x)g′ (x)

= L. (5.12)

Then

limx→a+

f (x)g(x)

= L. (5.13)

Here is a simple example which illustrates the use of this rule.

Example 5.17.3 Find limx→05x+sin3x

tan7x .

The conditions of L’Hôpital’s rule are satisfied because the numerator and denomina-tor both converge to 0 and the derivative of the denominator is nonzero for x close to 0.Therefore, if the limit of the quotient of the derivatives exists, it will equal the limit of theoriginal function. Thus,

limx→0

5x+ sin3xtan7x

= limx→0

5+3cos3x7sec2 (7x)

=87.

Sometimes you have to use L’Hôpital’s rule more than once.

Example 5.17.4 Find limx→0sinx−x

x3 .

Note that limx→0 (sinx− x) = 0 and limx→0 x3 = 0. Also, the derivative of the de-nominator is nonzero for x close to 0. Therefore, if limx→0

cosx−13x2 exists and equals L,

it will follow from L’Hôpital’s rule that the original limit exists and equals L. How-ever, limx→0 (cosx−1) = 0 and limx→0 3x2 = 0 so L’Hôpital’s rule can be applied againto consider limx→0

−sinx6x . From L’Hôpital’s rule, if this limit exists and equals L, it will

follow that limx→0cosx−1

3x2 = L and consequently limx→0sinx−x

x3 = L. But, limx→0−sinx

6x =

limx→0(−1

6

) sinxx = −1

6 . Therefore, by L’Hôpital’s rule, limx→0sinx−x

x3 = −16 .