# Limits and L'hopital's rule

Crystal037
Homework Statement:
lim x tends to 0 (((e^x+e^(-x)-2)/(x^2))^(1/x^2))
Relevant Equations:
lim x tends to 0 (e^x-1)/x=1
It is of the form (0/0)^infinity. I know how to solve 1^infinity form

## Answers and Replies

Mentor
2021 Award
Is it ##\lim_{x \to 0} \left( \dfrac{e^x+e^{-x}-2}{x^2}\right)^{\frac{1}{x^2}}\,?##

For L'Hôpital you will have do calculate the derivatives of numerator and denominator.

Crystal037
Yes but should i consider the power 1/x^2 while differentiating the numerators and denominator

Mentor
2021 Award
I would take it for both: ##\left( e^x+e^{-x} - 2\right)^{x^{-2}}## and ##\left( x^2 \right)^{x^{-2}}##. If I'm right, then the limit without the power of ##\frac{1}{x^2}## is ##1##, but with it, it isn't. But I'm not sure whether L'Hôpital will do. Maybe it needs more effort and a few Taylor expansions.

Mentor
I would take it for both: ##\left( e^x+e^{-x} - 2\right)^{x^{-2}}## and ##\left( x^2 \right)^{x^{-2}}##. If I'm right, then the limit without the power of ##\frac{1}{x^2}## is ##1##, but with it, it isn't. But I'm not sure whether L'Hôpital will do. Maybe it needs more effort and a few Taylor expansions.
I wouldn't do things this way. Instead, I would let ##y = \left(\frac{ e^x+e^{-x} - 2}{x^2}\right)^{1/x^2}##, take ln of both sides, and then take the limit. The result you get, if a limit exists will be the limit of the ln of that expression, so the final answer is obtained by exponentiating.

I haven't worked the problem, but this is the usual approach when dealing with limits with some expression raised to a variable power.

Mentor
2021 Award
Yes, but L'Hôpital was given in the thread title.

Staff Emeritus
Homework Helper
Yes, but L'Hôpital was given in the thread title.
The limits after taking the log can be evaluated using the Hospital rule. According to Mathematica, the limit is equal to ##e^{1/12}##.

You can simplify the algebra by using the fact that
$$\frac{e^x + e^{-x} - 2}{x^2} = \frac{(e^{x/2} - e^{-x/2})^2}{x^2} = \left[\frac{\sinh (x/2)}{x/2}\right]^2.$$ I had to apply L'Hopital's rule several times and eventually arrived at the correct result.

Mentor
Yes, but L'Hôpital was given in the thread title.
My response didn't preclude using L'Hopital's Rule in evaluating the limit.