# Limits and L'hopital's rule

• Crystal037
However, I would do it in a more simplified way by using the fact that$$\frac{e^x + e^{-x} - 2}{x^2} = \frac{(e^{x/2} - e^{-x/2})^2}{x^2}.$$In summary, L'Hôpital needs to calculate the derivatives of numerator and denominator. It may need more effort and a few Taylor expansions to do so. If it does, the final answer will be obtained by exponentiating.

#### 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

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

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.

fresh_42 said:
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.

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

fresh_42 said:
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.

fresh_42 said:
Yes, but L'Hôpital was given in the thread title.
My response didn't preclude using L'Hopital's Rule in evaluating the limit.