I couldn't find a way to get the answer for $$\lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}}$$
From my knowledge of L'Hopital's Rule, I see that this is some kind of $1^{\infty}$ indeterminate form since I know from previous results that $\lim\limits_{x \to 0} \left(\frac{\sin x}{x}\right)=1$. Proceeded to find the limit of its natural $\log$ which is $$\lim\limits_{x \to 0} \left( \frac{\ln(\frac{\sin x}{x})}{x^{2}}\right)$$ then got stuck when I got to $$\lim\limits_{x \to 0} \left( \frac{\cot x-\frac{1}{x}}{2x}\right)$$ as I now get $\infty/0$ and this hasn't happened to me before since I just started not long ago on this topic. Can someone give me a further hint as to which direction I should head to or recommend me another more suitable approach to solve this problem?
If it helps, the given answer is $e^{-1/6}$.