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!

$$\lim_{x\to 0} (\frac{\sin x}{x})^{\frac{1}{\tan^2 x}} =e^{\lim_{x\to 0}\frac{\sin x-x}{x\tan^2x}}= e^{\lim_{x\to 0}\frac{(\sin x-x)'}{(x^3)'}\frac{\tan^2x}{x^2}}=$$$$= e^{\lim_{x\to 0}\frac{\cos x-1}{3x^2}}= e^{\lim_{x\to 0}\frac{(\cos x-1)'}{(3x^2)'}}=e^{\lim_{x\to 0}\frac{-\sin x}{6x}}=e^{\frac{-1}{6}}. $$
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 $$\lim_{x→α}(f(x))^{g(x)}= \lim_{x→α}(1+f(x)−1)^{g(x)}= \lim_{x→α}[[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{(f(x)-1)}]^{g(x)}= \lim_{x→α}[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{\lim_{x→α}(f(x)-1)g(x)}= e^{\lim_{x→α}(f(x)-1)g(x)}.$$ 
