Prove $\lim_{x\to 0}\big(\frac{x}{\sin x}\big)^{1/x^2} = e^{\frac{1}{6}}$ As the title says, prove that $\lim_{x\to 0}\big(\frac{x}{\sin x}\big)^{1/x^2} = e^{\frac{1}{6}}$. I've tried using L'Hoptial's rule like so:
If $\lim_{x\to 0}\big(\frac{x}{\sin x}\big)^{1/x^2} = L$, then $\log L = \lim_{x\to 0}\frac{\log(\frac{x}{\sin x})}{x^2}$. Since this is of the form $\frac{0}{0}$, we can apply L'Hopital's rule to get $\log L=\lim_{x\to 0}\frac{\frac{1}{x}-\cot (x)}{2x}$. Since this is also of the form $\frac{0}{0}$, we can apply L'Hopital's rule again to get $\log L = \lim_{x\to 0}\frac{\csc^2(x)-\frac{1}{x^2}}{2}$.
I'm not sure where to go from here, or if I'm really headed in the right direction. I'm also not sure how to show that the intermediate steps are $\frac{0}{0}$, but this is less concerning to me. How do I show that this limit is $e^{\frac{1}{6}}$, or that the last limit is $\frac{1/3}{2} = \frac{1}{6}$?

After your first use of L'Hopital's Rule, do not immediately apply L'Hopital's Rule again, but do some algebra:
$$\frac{\frac1x-\cot x}{2x}=\frac{\sin x-x\cos x}{2x^2\sin x}\ .$$
Then continue with L'Hopital's Rule (and perhaps a bit more algebra at some stage).
