Calculate $\lim\limits_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}$ 
How to calculate 
  $$\lim_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}?$$

I know the result is $1/(6e)$.

Re-write the logarithm of the original expression as follows
\begin{eqnarray*}
\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}} &=&\frac{1}{x^{2}}\ln \left(
1+\left( \frac{\sin x}{x}-1\right) \right)  \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left( 
\frac{\sin x}{x}-1\right) }\frac{\left( \frac{\sin x}{x}-1\right) }{x^{2}} \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left( 
\frac{\sin x}{x}-1\right) }\frac{\left( \sin x-x\right) }{x^{3}}
\end{eqnarray*}
Now using the standard limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \ \lim_{u\rightarrow
0}\frac{\ln (1+u)}{u}=1,\ \ and\ \ \ \lim_{x\rightarrow 0}\frac{\sin x-x}{%
x^{3}}=-\frac{1}{6}.
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}}=1\cdot
\left( -\frac{1}{6}\right) =-\frac{1}{6}.
\end{equation*}
Taking back the exponential one gets
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sin x}{x}\right) ^{1/x^{2}}=e^{-1/6}.
\end{equation*}
