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!

$$\lim_{x \to 0} \left( \frac{\sin x}{x} \right) ^ {\cot^2{x}}$$
A simpler solution than l'Hopital's rule is to use Taylor expansions, as we are interested in the function as $x \to 0$. Note that $$\frac{\sin{x}}{x} = 1 - \frac{x^2}{6} + O(x^4)$$
Note that $\tan{x} = x + O(x^3)$, and thus $\cot^2{x} = 1/x^2 + O(x^{-4})$. Dropping higher terms we see $$\lim_{x \to 0} \left( \frac{\sin x}{x} \right) ^ {\cot^2{x}} = \lim_{x \to 0} \left(        1 - \frac{x^2}{6}\right)^{1/x^2} = \lim_{n \to \infty} \left( 1 - \frac{1}{6n}  \right)^n$$ which should look familiar.
