My Calculus teacher made up quiz off the top of his head
for a section on L'Hopital's Rule. The last question was to find the
limit as x approaches infinity in the equation cos(x)/ln(x). When you try
to use L'Hopital's you don't get an answer, yet when you graph the
equation, you see that it approaches 0. My teacher was not able to
explain why L'Hopital's rule falls flat on its face here. Can you tell us
how to solve the problem and why L'Hopital's rule does not work?
L'Hospital's Rule does work in this case. All the rule claims is that
lim f(x)/g(x) = lim f'(x)/g'(x)
which is true. Unfortunately, this information doesn't help you find
the limit. To do that in this case, the best tool is common sense:
cos(x) is bounded; ln(x) is not.
This is a great lesson whose relevance goes way beyond math. L'Hospital's
rule is not for finding limits. It's just a statement about functions and
their derivatives, and it doesn't care what you want to prove.
I believe that the rule is applicable when the denominator approaches zero.
In this case ln(x) goes to -infinity, as x approaches zero.
L'Hopital's rule is still applies as follows:
Lim(x -->0) of [cosx/lnx] = -lim(x -->0) of [(sinx)/(1/x)]
= -lim (x -->0) [x sinx] =0.
L'Hopital's rule is used in cases where both the numerator and the
denominator go to zero at some limits.
In the present case, for x -> infinity, L'Hopital's rule does not apply
Lim(x -->infinity) of [cosx/lnx] = ~ cosx/infinity =0. Cosine of x has
values between -1 to +1 and when divided by a large number approaches zero.
Dr. Ali Khounsary
Advanced Photon Source
Argonne National Laboratory
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Update: June 2012