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 $e^{-1/6}$.
Since $\lim_{x \to 0} \frac{\ln \left(\frac {\sin x}{x} \right)}{x^2}$ has the form $\frac 00$ you apply L'Hospital's rule and investigate whether $$ \lim_{x \to 0} \frac{ \frac{x}{\sin x} \frac{x \cos x - \sin x}{x^2}}{2x}$$ exists. But since $\lim_{x \to 0} \frac x{\sin x} = 1$ it suffices to investigate $$ \lim_{x \to 0} \frac{ \frac{x \cos x - \sin x}{x^2}}{2x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{2x^3}.$$ This has the form $\frac 00$ so apply L'Hospital again: $$\lim_{x \to 0}\frac{ -x \sin x + \cos x - \cos x}{6x^2} = \lim_{x \to 0} \frac{-\sin x}{6x} = -\frac 16.$$
To recap, you can conclude $$ \lim_{x \to 0} \frac{ \frac{x \cos x - \sin x}{x^2}}{2x} = -\frac 16$$ so that $$ \lim_{x \to 0} \frac{ \frac{x}{\sin x} \frac{x \cos x - \sin x}{x^2}}{2x} = - \frac 16$$ and thus $$\lim_{x \to 0} \frac{\ln \left(\frac {\sin x}{x} \right)}{x^2} = - \frac 16.$$