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.

Since everybody was 'clever', I thought I'd add a method that doesn't really require much thinking if you're used to asymptotics.
The power series for $\sin x$
$$\sin x = x + O(x^3)$$
We can compute the inverse of this power series without trouble. In great detail:
$$\begin{align}\frac{1}{\sin x} &= \frac{1}{x + O(x^3)}
\\ &= \frac{1}{x} \left( \frac{1}{1 - O(x^2))} \right)
\\ &= \frac{1}{x} \left(1 + O(x^2) \right)
\\ &= \frac{1}{x} + O(x)
\end{align}$$
going from the second line to the third line is just the geometric series formula. Anyways, now we can finish up:
$$\frac{1}{x} - \frac{1}{\sin x} = O(x)$$
$$ \lim_{x \to 0} \frac{1}{x} - \frac{1}{\sin x} = 0$$
If we wanted, we could get more precision: it's not hard to use the same method to show
$$ \frac{1}{\sin x} = \frac{1}{x} + \frac{x}{6} + O(x^3) $$
