calculus - Finding $lim_{n to infty} frac{1}{2^{n-1}}cot(frac{x}{2^n})$

Find: $$\lim_{n \to \infty} \frac{1}{2^{n-1}}\cot\left(\frac{x}{2^n}\right)$$

Can L' Hopital's rule be used to solve this? And differentiate it with respect to $x$ or $n$?

What I've found is that

\begin{equation}
\lim_{n \to \infty} \frac{1}{2^{n-1}}\cot\left(\frac{x}{2^n}\right) = \lim_{n \to \infty} \frac{\frac{1}{2^{n-1}}\cos\left(\frac{x}{2^n}\right)}{\sin\left(\frac{x}{2^n}\right)}
\end{equation}

which is of the form $\frac{0}{0}$, but I don't know how to go further from here. Any help is appreciated.

Answer

Hint:
$$
\lim_{x\to 0}\frac{\sin x}{x}=1
$$
and
$$
\lim_{x\to 0}\cos x =1.

$$