Question: > 
> $$\lim\_{x\to0}2^{\cot x}$$
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So it turns out that the limit does not exist but here's the thing. I can't compute this limit

$$\lim\_{x\to0}{\cot x}$$

this is all I did $$\lim\_{x\to0}{\cot x}=\lim\_{x\to0}\frac{\cos x}{\sin x}=\lim\_{x\to0}\frac{x\cos x }{x\sin x}=\lim\_{x\to0}\frac{\cos x}{x}$$

then nothing comes to mind. I feel like this is a really easy problem but I've been trying to solve thins for over an hour and I'm stuck. Could you please help me? Thanks in advance.

Answer: 
Since $\lim\_{x\to0^+}\frac1{\sin x}=\infty$, $\lim\_{x\to0^-}\frac1{\sin x}=-\infty$, and $\lim\_{x\to0}\cos x=1$, you have$$\lim\_{x\to0^+}\cot x=\infty\quad\text{and}\quad\lim\_{x\to0^-}\cot x=-\infty.$$So$$\lim\_{x\to0^+}2^{\cot x}=\infty\quad\text{and}\quad\lim\_{x\to0^-}2^{\cot x}=0.$$
Write $2^{\cot x}$ as $\exp(\cot x\log 2)$. Since the exponential function is continuous,
\begin{align}
\lim\_{x \to 0^+}\exp(\cot x\log2)&=\exp\left(\lim\_{x\to0^+}\cot x\log 2\right) \\
&= \exp\left(\infty\log 2\right) \\
&= \infty \, .
\end{align}
The same approach can be made for the left-hand limit, and the comparison of the two shows us that two-sided limit does not exist.