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.
 A: Hint:
$$
\lim_{x\to 0}\frac{\sin x}{x}=1
$$
and
$$
\lim_{x\to 0}\cos x =1.
$$
A: 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*}
A: Try this- We know that $\lim_{x \to \ 0}$$ \text{tan}(x)\over x$$ = 1$ Now, $\text{cot}(\frac{x}{2^n})$= $2^n\over x$ , Then the answer is $2\over x$. You can use the L'Hopital's rule but it just gets harder.
A: It is so much simpler to rewrite it as $\;2^{n-1}\dfrac{1}{\tan\dfrac{x}{2^n}}$, and use equivalents:
$$\tan u\sim_0 u,\quad\text{hence}\quad \frac{1}{2^{n-1}}\cot\Bigl(\frac{x}{2^n}\Bigr) \sim_\infty \frac{1}{2^{n-1}}\frac{1}{\dfrac{x}{2^n}}=\frac 2x.$$
A: You can as well compute
$$
\lim_{t\to\infty}\frac{2}{t}\cot\frac{x}{t}
$$
because if this limit exists, also your sequence converges at the same limit as $\lim_{n\to\infty}2^n=\infty$.
Now you can use substitutions more freely: use $u=x/t$, so you get (for $x>0$)
$$
\lim_{t\to0^+}\frac{2}{x}\frac{u}{\tan u}=\frac{2}{x}
$$
For $x<0$ the limit is for $t\to0^-$, but the final value is the same.
