$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$ Could anyone give me hints on how to solve this limit without L'Hospital rule?
$$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$$
I tried using standard formualaes but got nowhere.
 A: By Taylor, $e^x+e^{-x}=2+x^2+\frac{x^4}{12}+o(x^4)$, hence
$$\lim_{x\to 0}\left({e^x+e^{-x}-2\over x^2}\right)^{1/x^2}=\lim_{x\to0}\left(1+\frac{x^2}{12}+o(x^2)\right)^{1/x^2}=\lim_{t\to0}(1+t+o(t))^{1/12t}=e^{1/12}.$$
A: If $L$ is the desired limit then $$\begin{aligned}\log L &= \log\left\{\lim_{x \to 0}\left(\frac{e^{x} + e^{-x} - 2}{x^{2}}\right)^{1/x^{2}}\right\}\\
&= \lim_{x \to 0}\log\left(\frac{e^{x} + e^{-x} - 2}{x^{2}}\right)^{1/x^{2}}\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{2\cosh x - 2}{x^{2}}\right)\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\log\left(1 + \frac{2\cosh x - 2 - x^{2}}{x^{2}}\right)\\
&= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\dfrac{\log\left(1 + \dfrac{2\cosh x - 2 - x^{2}}{x^{2}}\right)}{\dfrac{2\cosh x - 2 - x^{2}}{x^{2}}}\cdot\dfrac{2\cosh x - 2 - x^{2}}{x^{2}}\\
&= \lim_{t \to 0}\frac{\log(1 + t)}{t}\cdot\lim_{x \to 0}\frac{2\cosh x - 2 - x^{2}}{x^{4}}\\
&= \lim_{x \to 0}\frac{2\cosh x - 2 - x^{2}}{x^{4}}\\
&= \lim_{x \to 0}\frac{1}{x^{4}}\left\{2\left(1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \cdots\right) - 2 - x^{2}\right\}\\
&= \frac{1}{12}\end{aligned}$$ hence $L = e^{1/12}$. In the above derivation we have used the substitution $$\begin{aligned}t &= \frac{2\cosh x - 2 - x^{2}}{x^{2}}\\
&= \frac{1}{x^{2}}\left\{2\left(1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \cdots\right) - 2 - x^{2}\right\}\\
&= \frac{x^{2}}{12} + \cdots\end{aligned}$$ and clearly $t \to 0$ as $x \to 0$. I am not sure if we can do this limit problem without the use of Taylor's series or L'Hospital's rule.
