Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\begin{align}
& \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\
& \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\
& e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\
& e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\
& e^{-\infty} = 0\\
\end{align}
$$

Edit #1:
Continuing after the mistake of the $\tan(x)$:
$$\begin{align}
 & e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\
\end{align}
$$
 A: Take the logarithm of your limit then use L'Hopital and Taylor series. Let the expression in the limit be $L$:
$$\begin{align}
\ln L&=\ln\left[ \left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \right] \\[2ex]
 &=  \frac{\ln\left(\frac{\sin(x)}{x}\right)}{x^2} \\[2ex]
 &\to\frac{\frac{x}{\sin x}\cdot \frac{x\cos x-\sin x}{x^2}}{2x}\qquad \text{(L'Hopital)}\\[2ex]
 &=  \frac{x\cos x-\sin x}{2x^2\sin x} \\[2ex]
 &\to\frac{x(1-\frac{x^2}2)-(x-\frac{x^3}6)}{2x^2(x)}\qquad \text{(Taylor series)} \\[2ex]
 &=  \frac{-\frac 13x^3}{2x^3} \\[2ex]
 &=  -\frac 16
\end{align}$$
Therefore $L\to e^{-1/6}$. (I left a few minor gaps for you to fill in.)
If you don't like Taylor series, we can finish the whole thing with L'Hopital, though this way takes more steps.
$$\begin{align}
\ln L&\to\frac{x\cos x-\sin x}{2x^2\sin x} \qquad\text{(from above, before Taylor series)}\\[2ex]
 &\to\frac{x\cdot -\sin x +\cos x-\cos x}{2x^2\cos x + 4x\sin x} \qquad\text{(L'Hopital)}\\[2ex]
 &=  \frac{-\sin x}{2x\cos x + 4\sin x} \\[2ex]
 &\to\frac{-\cos x}{2x\cdot -\sin x+2\cos x+4\cos x}  \qquad\text{(L'Hopital)}\\[2ex]
 &=  \frac{-\cos x}{-2x\sin x+6\cos x} \\[2ex]
 &\to\frac{-1}{0+6} \\[2ex]
 &=  -\frac 16
\end{align}$$
A: Using Taylor series: 
$$\frac{\sin(x)}{x}=1- \frac{1}{3!}x^2+o(x^2).$$
So using this formula we get:
$$
\ln \left( \frac{\sin(x)}{x}\right)=\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right).
$$
Now we can use the Taylor expansion for the $\ln$ function so we get:
$$
\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right)= - \frac{1}{3!}x^2+o(x^2).
$$
Now divide all by $x^2$ and compute the limit while $x$ goes to zero:
$$
\frac{- \frac{1}{3!}x^2+o(x^2)}{x^2}
$$
as $x\rightarrow 0$, we obtain that
$$
\lim_{x \rightarrow 0} \ln \left( \frac{\sin(x)}{x}\right)=
-\frac{1}{6}.
$$
At the end
$$
\lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x} \right)^{\frac{1}{x^2}}=
e^{\lim_{x \rightarrow 0}\frac{\ln\frac{\sin(x)}{x}}{x^2}}=
e^{-\frac{1}{6}}.
$$
A: :(  I have not enough reputation to even add a comment...
Yes, the cot<-->tan problem. 
Comment: it should be equal to $e^{-\frac{1}{6}}$, if the cot<-->tan was not there.
A: \begin{align*}
\left(\frac{\sin x}x\right)^{\frac1{x^2}}
&=\left[1+\left(\frac{\sin x}x-1\right)\right]^{\frac1{x^2}}\\
&=\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{\frac1{x^2}}\\
&=\left[\underbrace{\left[1+\left(\frac1{x/(\sin x-x)}\right)\right]^{[x/(\sin x-x)]}}_{=:A(x)}\right]^{\frac{\sin x-x}{x^3}}
\end{align*}
Next observe that
$\bullet\sin x=x-\frac{x^3}{6}+o(x^3) \Longrightarrow\frac{\sin x-x}{x^3}\stackrel{x\to0}{\longrightarrow}-\frac16$;
$\bullet\lim_{\xi\to\infty}\left(1+\frac1\xi\right)^{\xi}=e \Longrightarrow A(x)\stackrel{x\to0}{\longrightarrow} e \;\;(\mbox{in fact} \;\;x/(\sin x-x)\stackrel{x\to0}{\longrightarrow}\infty)$
thus we immediately get
$$
\lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=e^{-\frac16}
$$
