로피탈 정리

$\displaystyle \lim_{x \to 0} \cos x =1$ and $\displaystyle \lim_{x \to 0}\frac{1}{x^2}=\infty$, so $\displaystyle \lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}$ has the indeterminate form $1^\infty$. We want to use l'Hopital, but we need to get our function into a fractional indeterminate form. So, we'll use a logarithm.

$$\begin{aligned} y:&=(\cos x)^{\frac{1}{x^2}}\\ \log y &= \log \left((\cos x)^{\frac{1}{x^2}}\right)=\frac{1}{x^2}\log(\cos x)=\frac{\log \cos x}{x^2}\\ \lim_{x \to 0}\log y &=\lim_{x \to 0}\underbrace{\frac{\log \cos x}{x^2}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} = \lim_{x \to 0} \frac{\frac{-\sin x}{\cos x}}{2x}= \lim_{x \to 0} \underbrace{\frac{-\tan x}{2x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} = \lim_{x \to 0}\frac{-\sec^2x}{2}\\ &=\lim_{x \to 0}\frac{-1}{2\cos^2 x}=-\frac{1}{2}\\ \text{Therefore, } \lim_{x \to 0} y &=\lim_{x \to 0}e^{\log y}=e^{-1/2}=\frac{1}{\sqrt{e}} \end{aligned}$$