Evaluate $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$ I need to find the following limit: $$\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$
I started this way: $$\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \ln\left[\frac {\sin x}{x}\right]}$$
So it's enough to find: $$\lim_{x\to 0^+}\frac{\ln(\frac {\sin x}{x})}{x}$$
I tried to use L'Hôpital's rule but it got me nowhere. Any ideas?

There is an interesting massive overkill: we may prove that in a right neighbourhood of the origin we have
$$1-\frac{x^2}{6}\leq\frac{\sin x}{x}\leq \exp\left(-\frac{x^2}{6}\right) \tag{1}$$
hence the wanted limit is trivially $\exp(0)=1$. By the Weierstrass product for the sine function:
$$ \log\left(\frac{\sin x}{x}\right) = \sum_{n\geq 1}\log\left(1-\frac{x^2}{\pi^2 n^2}\right)\leq\sum_{n\geq 1}-\frac{x^2}{\pi^2 n^2}=-\frac{\zeta(2)}{\pi^2}x^2=-\frac{x^2}{6}\tag{2} $$
and we are done, since the inequality on the left of $(1)$ just follows from the Taylor series of $\text{sinc}(x)$.
