What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? Find the limit: 
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$
I am not able to find it because I don't know how to prove or disprove $0$ is the answer.

Simplify to have $$\frac{\sin x-x }{x\sin x}$$ and consider Maclaurin's series for $$\sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}-...$$ 
So you have $$\frac{(x-\frac {x^3}{3!}+\frac {x^5}{5!}-...)-x}{x(x-\frac {x^3}{3!}+\frac {x^5}{5!}+...)}=\frac{(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}.$$ 
Finding the limit as $x\rightarrow 0$, we have;
$$\frac{\lim_{x\rightarrow 0}(-\frac {x}{3!}+\frac {x^3}{5!}-...)}{\lim_{x\rightarrow 0}(1-\frac {x^2}{3!}+\frac {x^4}{5!}-...)}=\frac{0}{1}=0.$$
which is the required answer.
