$\lim_{x \to 0} \left ({e^x+e^{-x}-2\over x^2} \right )^{1\over x^2}$

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.

• Did you try taking the logarithm first? – Harald Hanche-Olsen Sep 8 '14 at 8:18

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.
• Why the limit under log is $1$? the fraction associated with $\cosh x$ is not going to $0$ right? it is $0$ by $0$ form – Marso Sep 9 '14 at 3:40
By Taylor, $e^x+e^{-x}=2+x^2+\frac{x^4}{12}+o(x^4)$, hence $$\lim_{x\to 0}\left({e^x+e^{-x}-2\over x^2}\right)^{1/x^2}=\lim_{x\to0}\left(1+\frac{x^2}{12}+o(x^2)\right)^{1/x^2}=\lim_{t\to0}(1+t+o(t))^{1/12t}=e^{1/12}.$$