Q:

L'Hopital rule to solve $ \lim\limits_{x\to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} $

I am trying to solve the following limit:
\begin{equation}
\lim_{x\to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}}
\end{equation}
I tried using $ \ln $ to get a exponential expression for the equation as follows:
\begin{equation}
\text{Let} \,\, y = \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}}
\end{equation}
Apply $ \ln $ to bring down power and apply exponential:
\begin{equation}
y = \exp\left(\frac{1}{x^{2}}\ln\left(\frac{\sin x}{x}\right)\right)
\end{equation}
I have no idea how to solve the limit for $y$. From what I read, it can be done using l'Hopital  rule but I am unable to get an indeterminate form no matter how I try. Could somone please assist me?

A:

Marty Cohen's answer gives the most elegant solution, but since you asked for a solution using L'Hopital's rule, you should use (several applications of) it on the power:
$$\lim_{x\to0}\frac{\ln(\sin x/x)}{x^2}=\lim_{x\to0}\frac{\frac{x}{\sin x}\cdot\frac{x\cos x-\sin x}{x^2}}{2x}=\lim_{x\to0}\frac{x\cos x-\sin x}{2x^2\sin x}=\lim_{x\to0}\frac{-\sin x}{2(2\sin x+x\cos x)}$$
All of these steps are justified because we get $0/0$ each time. Iterating one last time, we get
$$\lim_{x\to0}\frac{-\cos x}{2(3\cos x-x\sin x)}$$
At last, this limit does not yield an indeterminate form. Can you take it from here?

