Hairy Limit With L'Hopital's Rule

by Saoni
(Singapore)

This is a very hairy and challenging limit problem posted by Saoni from Singapore. This is a problem where you need to apply L'Hopital's rule. However, the problem is not initially an indetermination of the type 0/0 or infinity/infinity. It is of the
type 1 to (infinity):

To transform this into 0/0 or infinity/infinity we apply natural logarithm. That is, suppose that we define the function:



If we apply natural logarithms to both sides:



Now, as the argument of the logarithm approaches 1, the natural logarithm approaches zero. The denominator obviously approaches zero when x approaches zero. So, if we take the limit as x approaches zero of this new function, we have 0/0. Therefore, we can apply L'Hopital's rule to this limit.

To solve this you need to apply L'Hopital's twice, and it is not an easy problem even from here. To get the original limit back, we must use a property of continuous functions, we can first take the limit and then evaluate the function:



If you solve the limit to the left, you'll have the natural logarithm of the limit we originally were trying to solve. Then, we just need to solve the logarithmic equation.

I won't post the step by step solution of this problem, as I want you to have fun also. As I said, you'll need to apply L'Hopital's rule twice and the final answer is:



Have fun!

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