<s>[INST] Find $\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2x}}$ Is there any other way to compute the problem $$\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2(x)}}$$ 
I try to use L'Hospital rule but it's too complicated.
 [/INST] In the same spirit as other answers but assuming that you want to go beyond the limit.
$$A=\left(\frac {\sin (x)}{x}\right)^{\frac{1}{\sin^2(x)}}\implies \log(A)={\frac{1}{\sin^2(x)}}\log\left(\frac {\sin (x)}{x}\right)$$ Using Taylor for one more term
$$\frac {\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\left(\frac {\sin (x)}{x}\right)=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$
$$\log(A)=\frac{-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right) } { x^2-\frac{x^4}{3}+O\left(x^6\right)}=\frac{-\frac{1}{6}-\frac{x^2}{180}+O\left(x^4\right) } { 1-\frac{x^2}{3}+O\left(x^4\right)}$$ Now, long division 
$$\log(A)=-\frac{1}{6}-\frac{11 x^2}{180}+O\left(x^4\right)$$ Finally, Taylor again 
$$A=e^{\log(A)}=\frac{1}{\sqrt[6]{e}}-\frac{11 x^2}{180 \sqrt[6]{e}}+O\left(x^4\right)$$
This is even a good approximation : try $x=\frac \pi 6$ (quite far away from $0$). The exact formula would give $$A=\frac{81}{\pi ^4}\approx 0.831545$$ while the above approximation would give $$\frac{1}{\sqrt[6]{e}}\left(1-\frac{11 \pi ^2}{6480}\right)\approx 0.832300$$
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