Evaluate analytically...$\lim\limits_{x\to 0} \left(\frac{\sin(x)}{x}\right)^{\cot^2(x)}$ Can someone help?  My professor likes to call these easy problems...Wolfram Alpha says the answer should be $e^{-1/6}$.  Every time I do it I get something different.  Help!

It might help to notice first that $\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1$ and $\cot(x) \rightarrow \infty$ . So, we have to invoke L'Hospital's rule. The standard way to do this is:
Let $y=(\frac{\sin(x)}{x})^{\cot(x)^2}$. Then,
$\ln(y)=\cot(x)^2 \ln(\frac{\sin(x)}{x})$ which is of the form $\infty$ times $0$. So,
we manipulate this to get:
$\lim_{x \rightarrow 0}\ln(\frac{\sin(x)}{x})/\tan(x)^2$ which is of the form $0/0$. Now use L'Hospital's rule to get $\lim_{x \rightarrow 0}\ln(y)$. Then the final answer will be $e^{\lim_{x \rightarrow 0}\ln(y)}$.
Try working out the details. Since you haven't given any indication of what you have tried, I shall not put the rest of the work.
