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.

Let $\displaystyle \varepsilon=\frac{x}{2^{n}}$ and applying L'Hospital's rule,
\begin{align*}
  \lim_{n\to \infty} \frac{\cot \frac{x}{2^{n}}}{2^{n-1}}  &=
  \lim_{\varepsilon \to 0} \frac{2\varepsilon}{x\tan \varepsilon} \\ &=
  \lim_{\varepsilon \to \infty} \frac{2}{x\sec^{2} \varepsilon} \\ &=
  \frac{2}{x}
\end{align*}
