# Limits and L'hopital's rule

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

fresh_42
Mentor
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.

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

fresh_42
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.

Mark44
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.

fresh_42
Mentor
Yes, but L'Hôpital was given in the thread title.

vela
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.

Mark44
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.