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.
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\ln L$
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\lim\limits_{x \to 0}\frac{\ln\frac{\sin x}{x}}{\tan ^2x}=\lim\limits_{x \to 0}\frac{x}{2\sin x\tan x \sec^2x}$

Composing Taylor series one piece at the time
$$y=\Bigg[\frac{\sin (x)}{x}\Bigg]^{\cot ^2(x)} \implies \log(y)={\cot ^2(x)}\log\Bigg[\frac{\sin (x)}{x}\Bigg]$$
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\Bigg[\frac{\sin (x)}{x}\Bigg]=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
$$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2 x^5}{945}+O\left(x^6\right)$$
$$\cot^2(x)=\frac{1}{x^2}-\frac{2}{3}+\frac{x^2}{15}+\frac{2 x^4}{189}+O\left(x^5\right)$$
$$\log(y)=-\frac{1}{6}+\frac{19 x^2}{180}-\frac{22 x^4}{2835}+O\left(x^6\right)$$
$$y=e^{\log(y)}=\frac{1}{\sqrt[6]{e}}+\frac{19 x^2}{180 \sqrt[6]{e}}+O\left(x^4\right)$$
