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.

METHOD I
Firstly, notice that the expression under the limit is an odd function and consider that $\sin(x)<x$. Then we have that:
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\frac{\sin x - x}{x\sin x}\le\lim_{x \rightarrow 0}\frac{\sin x - x}{x^2}\le\lim_{x \rightarrow 0}\frac{\tan x - x}{x^2}=0$$
As regards the last limit you wanna see my proof here.
Q.E.D.
METHOD II
$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)= \lim_{x \rightarrow 0}\frac{\sin x - x}{x\sin x}\le\lim_{x \rightarrow 0}\frac{\sin x - x}{x^2}=\lim_{x \rightarrow 0}x\cdot\frac{\sin x - x}{x^3}=0\cdot-\frac{1}{6}=0$$
Let's solve now the auxiliary limit I used (elementarily):
$$L=\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}=\lim_{x \rightarrow 0}\frac{\sin 2x - 2x}{8x^3}=\lim_{x \rightarrow 0}\frac{\sin x \cos x  - x}{4x^3}=\lim_{x \rightarrow 0}\frac{\sin x \cos x  -x\cos x + x\cos x- x}{4x^3}=\lim_{x \rightarrow 0}\frac{\cos x(\sin x \  -x) }{4x^3}-\lim_{x \rightarrow 0}\frac{(1 - \cos x) }{4x^2}=$$ 
$$\lim_{x \rightarrow 0} \cos x \cdot\frac{L}{4} -\frac{1}{8}=\frac{L}{4}-\frac{1}{8}$$
$$L=\frac{L}{4}-\frac{1}{8}$$
$$L=-\frac{1}{6}.$$
Q.E.D.
