Calculate $\lim\limits_{x \to 0}(\frac{\sin x}{x})^{\cot ^2x}$ 
Calculate
$$\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}$$

I tried to use the natural logarithm and L'hopital rule but it didn't help me.
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\ln L$
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\lim\limits_{x \to 0}\frac{\ln\frac{\sin x}{x}}{\tan ^2x}=\lim\limits_{x \to 0}\frac{x}{2\sin x\tan x \sec^2x}$
 A: Partial solution (fill in details): Note that $$\left(\frac{\sin x}{x}\right)^{\cot ^2x}=e^{\ln \left(\frac{\sin x}{x} \right)^{\cot^{2}(x)}}=e^{\cot^{2}(x)\ln\left( \frac{\sin x}{x} \right)}$$
and since that $$\cot(x)=\frac{\cos(x)}{\sin(x)} \implies \cot^{2}(x)=\frac{\cos^{2}(x)}{\sin^{2}(x)}$$
By, L'Hópital theorem we know that $$\lim_{x\to 0} \frac{\ln\left(\frac{\sin(x)}{x} \right)}{\sin^{2}(x)}=\lim_{x\to 0} \frac{x\cos(x)-\sin(x)}{2x\cos(x)\sin^{2}(x)}$$
and we know that $$\lim_{x\to 0} \frac{\sin(x)}{x}=1$$
So, we can see that $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}=e^{\lim_{x\to 0}\cot^{2}(x)\ln\left( \frac{\sin x}{x} \right)}=e^{-\frac{1}{6}}=\boxed{\frac{1}{\sqrt[6]{e}}}$$
A: If $$L=\lim_{x\to a} f(x)^{g(x)} \to 1^{\infty}$$,
then $$L=\exp[\lim_{x\to a} g(x)(f(x)-1)]$$
So here
$$L=\exp[\lim_{x \to 0} \cot^2 x(\frac{\sin x}{x}-1)]=\exp[\frac{(\sin x-x)\cos^2x}{\sin^2 x}].$$
$$L=\lim_{x \to 0} \exp\frac{\sin x -x}{x}\frac{1+\cos 2x}{1-\cos 2x}=\lim_{x \to 0} \exp\frac{-x^3/6+...}{x}\frac{2-2x^2+...}{2x^2+..}=e^{-\frac{1}{6}}.$$
We have used $\sin z \to z-z^3/6+...$ and $\cos z=1-z^2/2+..$ When $|z|$ is very small.
A: Here's a direct way with Taylor Series:
$$
\begin{aligned}\lim _{x\rightarrow 0}\frac{\ln \frac{\sin x}{x}}{\tan ^2x}
=\lim _{x\rightarrow 0}\frac{-\frac{x^2}{6}+o\left( x^4 \right)}{x^2+o\left( x^4 \right)}=-\frac{1}{6}\end{aligned}
$$
Thus
$$
\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x} = e^{-\frac16}
$$
A: Composing Taylor series one piece at the time
$$y=\Bigg[\frac{\sin (x)}{x}\Bigg]^{\cot ^2(x)} \implies \log(y)={\cot ^2(x)}\log\Bigg[\frac{\sin (x)}{x}\Bigg]$$
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\Bigg[\frac{\sin (x)}{x}\Bigg]=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
$$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2 x^5}{945}+O\left(x^6\right)$$
$$\cot^2(x)=\frac{1}{x^2}-\frac{2}{3}+\frac{x^2}{15}+\frac{2 x^4}{189}+O\left(x^5\right)$$
$$\log(y)=-\frac{1}{6}+\frac{19 x^2}{180}-\frac{22 x^4}{2835}+O\left(x^6\right)$$
$$y=e^{\log(y)}=\frac{1}{\sqrt[6]{e}}+\frac{19 x^2}{180 \sqrt[6]{e}}+O\left(x^4\right)$$
