On-line Math 21

4.4  l'Hôpital's Rule

Theorem 1 Let f and g be differentiable on some interval, with a in that interval, and assume they satisfy

lim x® a  f(x) = 0,  and   lim x® a  g(x) = 0.

Then,

lim x® a  f(x) g(x) = lim x® a  f¢(x) g¢(x) .

Proof. This result uses the Generalized MVT from earlier in this chapter. Let's deal with limits on the right. Assume that

lim x® a+  f(x) = 0,  and   lim x® a+  g(x) = 0,

and that f and g are differentiable on some interval (a,b). If you set g(a) = f(a) = 0 , then the functions f and g will be continuous on [a,b] and differentiable on (a,b) , so will fit the hypotheses of the generalized MVT. Now, there should be some question about the hypothesis that g¢(x) ¹ 0 , but certainly the limit

lim x® a+  f¢(x) g¢(x)

won't make sense unless g¢(x) ¹ 0 on some interval near a , so we can choose b so that g¢(x) ¹ 0 on (a,b) . Then, by the generalized MVT, if x Î (a,b) ,

f(x)-f(a) g(x)-g(a) = f(x) g(x) = f¢(c) g¢(c) ,

for some c between a and x . Since c is trapped between a and x , as x approaches a , c will also approach a , and so

lim x® a+  f(x) g(x) = lim c® a+  f¢(c) g¢(c) = lim x® a+  f¢(x) g¢(x) .

Since the same idesa work out for the limits from the left, the statement follows.

Copyright (c) 2000 by David L. Johnson.