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!
 A: $$\lim_{x \to 0} \left( \frac{\sin x}{x} \right) ^ {\cot^2{x}}$$
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.
A: 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.
A: $$\lim_{x\to 0} (\frac{\sin x}{x})^{\frac{1}{\tan^2 x}} =e^{\lim_{x\to 0}\frac{\sin x-x}{x\tan^2x}}= e^{\lim_{x\to 0}\frac{(\sin x-x)'}{(x^3)'}\frac{\tan^2x}{x^2}}=$$$$= e^{\lim_{x\to 0}\frac{\cos x-1}{3x^2}}= e^{\lim_{x\to 0}\frac{(\cos x-1)'}{(3x^2)'}}=e^{\lim_{x\to 0}\frac{-\sin x}{6x}}=e^{\frac{-1}{6}}. $$
I used a "shortcut" for $1^{\infty}$ and applied the rule of L'Hospital twice.
"Shortcut": 
If 
$\lim_{x→α}f(x)=1$ and $\lim_{x→α}g(x)= \infty $ 
then $$\lim_{x→α}(f(x))^{g(x)}= \lim_{x→α}(1+f(x)−1)^{g(x)}= \lim_{x→α}[[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{(f(x)-1)}]^{g(x)}= \lim_{x→α}[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{\lim_{x→α}(f(x)-1)g(x)}= e^{\lim_{x→α}(f(x)-1)g(x)}.$$ 
A: Use $$ \lim_{x → 0}\left(1+x\right)^\frac{1}{x}=e $$ In General
$$ \lim_{x → 0}\left(1+f\left(x\right)\right)^\frac{1}{f\left(x\right)}=e $$ for $$ \lim_{x → 0}f\left(x\right)=0$$ Now
$$\lim_{x → 0} \frac{\sin x-x}{x}=0$$ So
$$L=\lim_{x → 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}=\lim_{x → 0}\left(1+\frac{\sin x-x}{x}\right)^{\cot ^2x}=\lim_{x → 0}\left(1+\frac{\sin x-x}{x}\right)^{\left(\frac{x}{\sin x-x}\right)\left(\frac{\sin \left(x\right)-x}{x\tan^2x}\right)}$$
$$L=e^{\lim_{x → 0}\left( \frac{\sin x-x}{x\tan ^2x}\right)}=e^{\lim_{x → 0} \frac{\sin x-x}{x^3}}=e^\frac{-1}{6}$$
