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.
 A: HINT: Take logs and use the second order Taylor polynomials of $\sin^2x$ and $\ln\big(\frac{\sin x}x\big)$.
A: $$\frac{\sin x}x=1-\frac{x^2}{6}+O(x^4)$$
$$\ln\frac{\sin x}x=-\frac{x^2}{6}+O(x^4)$$
$$\ln\left(\frac{\sin x}x\right)^{1/\sin^2x}=-\frac{x^2}{6\sin^2x}+O(x^4/\sin^2 x)\to-1/6$$
as $x\to0$ etc.
A: I think the  L'Hospital rule helps here very well:
$$\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac{1}{\sin^2x}}=\lim_{x \to 0} \left(\frac {1+\sin x}{x}-1\right)^{\frac{1}{\frac{\sin x}{x}-1}\cdot\frac{\frac{\sin x}{x}-1}{\sin^2x}}=$$
$$=e^{\lim\limits_{x\rightarrow0}\frac{\sin{x}-x}{x\sin^2x}}=e^{\lim\limits_{x\rightarrow0}\left(\frac{\sin{x}-x}{x^3}\cdot\frac{x^2}{\sin^2x}\right)}=e^{\lim\limits_{x\rightarrow0}\frac{\sin{x}-x}{x^3}}=$$
$$=e^{\lim\limits_{x\rightarrow0}\frac{\cos{x}-1}{3x^2}}=e^{-\frac{1}{6}\lim\limits_{x\rightarrow0}\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}}=\frac{1}{\sqrt[6]e}$$
A: 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 $$\frac{1}{\sqrt[6]{e}}\left(1-\frac{11 \pi ^2}{6480}\right)\approx 0.832300$$
A: $$\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac{1}{\sin^2x}}$$
$$=\left(\lim_{x \to 0} \left(1+\dfrac{\sin x-x}x\right)^{\dfrac x{\sin x-x}}\right)^{\lim_{x \to 0} \dfrac{\sin x-x}{x^3}\cdot\left(\lim_{x \to 0}\dfrac x{\sin x}\right)^2}$$
Now by Are all limits solvable without L'Hôpital Rule or Series Expansion, $$\lim_{x \to 0}\dfrac{\sin x-x}{x^3}=\dfrac16$$
$\implies \lim_{x \to 0}\dfrac{\sin x-x}x=0$
Consequently, $$\lim_{x \to 0}\left(1+\dfrac{\sin x-x}x\right)^{\dfrac x{\sin x-x}}=e$$
Can you take it from here?
