$$\lim_{x\to0}2^{\cot x}$$
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.
Best 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$$\lim_{x\to0^+}2^{\cot x}=\infty\quad\text{and}\quad\lim_{x\to0^-}2^{\cot x}=0.$$