How do I solve this limit: $\lim _{x \to 0} \left(\frac{ \sin x}{x}\right)^{1/x}$? $$\lim _{x \rightarrow 0} \left(\frac{
\sin x}{x}\right)^{1/x}$$
I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.

Notice by Young's Inequality since $(\frac{1}{x} + (1 - \frac{1}{x})) = 1$, then
$$ (\frac{\sin x}{x} )^{1/x}=(\frac{\sin x}{x} )^{1/x} 1^{1 - \frac{1}{x}} \leq \frac{\sin x}{x^2} + 1 - \frac{1}{x} = \frac{\sin x - x}{x^2} + 1$$
Now, for positive $x$ and for $a \leq 1$, we have that $a^x \leq x +1 $. Hence $a^{1/x} \geq \frac{1}{x+1}$. Now, since $\frac{\sin x}{x} \leq 1$, we apply this inequality with $a = \frac{\sin x}{x} $ to obtain
$$  (\frac{\sin x}{x} )^{1/x} \geq \frac{1}{x+1}$$. Hence we have
$$  \frac{1}{x+1} \leq (\frac{\sin x}{x} )^{1/x} \leq \frac{\sin x - x}{x^2} + 1$$.
Now, since $\lim{ \frac{1}{x+1} } = 1 $ and 
$$ \lim (\frac{\sin x - x}{x^2} + 1 ) = \lim ( \frac{\sin x - x}{x^2} ) = 1 + \lim ( \frac{\cos x - 1}{2x} ) = 1 + \lim ( \frac{- \sin x}{2} ) = 1 $$
Now, result follows by the squeeze trick.
