Find the value of $\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$. I am stuck with the following problem :

Find the value of $$\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$$

My try : Let $$p=\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}\implies \log p=\lim_{x \to 0}{\frac {1}{x^2}}\log\left(\frac {\sin x}{x}\right)$$...After applying l'hospitals rule few times  I get 
$$\log p= \lim_{x \to 0}\frac{1-\sec^2x}{4x \tan x+2x^2\sec^2x}$$.. we can again apply l'hospitals rule ,but the calculations get bigger and bigger..Is there any other easier way around or I am missing something?
 A: From the Taylor expansion of $\sin x$ around $0$, namely $\sin u = u -\frac{u^3}{6} + o(u^3)$; and that of $\ln(1+u)$, specifically $\ln(1+u)=u+o(u),$ we get
$$\begin{align}
\left( \frac{\sin x}{x}\right)^{1/x^2}
&= \left(1-\frac{x^2}{6} + o(x^2)\right)^{1/x^2}
= e^{\frac{1}{x^2}\ln\left(1-\frac{x^2}{6} + o(x^2)\right) }
= e^{\frac{1}{x^2}\left(-\frac{x^2}{6} + o(x^2)\right) }\\
&= e^{-\frac{1}{6} + o(1)}
\xrightarrow[x\to0]{}\boxed{e^{-\frac{1}{6}}}
\end{align}$$
A: Applying L'Hospital's rule rwice we will get

$$\lim _{ x\to 0 } \left( \frac { \sin  x }{ x }  \right) ^{ \frac { 1 }{ x^{ 2 } }  }=\lim _{ x\to 0 } \left[ { \left( 1+\frac { \sin  x-x }{ x }  \right)  }^{ \frac { x }{ \sin { x-x }  }  } \right] ^{ \frac { \sin { x-x }  }{ x^{ 3 } }  }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \sin { x-x }  }{ x^{ 3 } }  }  }=\\ \overset { L'hospital's }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3x^{ 2 } }  }  }\overset { \quad L'hospital's }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\sin { x }  }{ 6x }  }  }=\color{red}{{ e }^{ -\frac { 1 }{ 6 }  }}$$

A: You can do this with L'Hopital, if you massage the expression you get from the first round a bit:
$${f(x)\over g(x)}={\log\sin x-\log x\over x^2}\implies{f'(x)\over g'(x)}={{\cos x\over\sin x}-{1\over x}\over2x}={x\cos x-\sin x\over2x^2\sin x}$$
At worst, another three rounds of L'Hopital will leave something nonzero in the denominator.  You can speed things up by noting that
$${(x\cos x-\sin x)'\over(2x^2\sin x)'}={-x\sin x\over4x\sin x+2x^2\cos x}={-1\over4+2{x\over\sin x}\cos x}\to{-1\over4+2\cdot1\cdot1}={-1\over6}$$
