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.
For fun, and because of the pre-calculus tag, we give a proof without calculus. It turns out that there is a geometric argument that $|x-\sin x|$ is less than a constant times $|x^3|$ for $x$ near $0$.    

I will need some help from you, to draw the missing picture.  We have
$$\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x}.$$
Let
$$f(x)=\frac{x-\sin x}{x\sin x}$$
(the change of sign is for convenience). We will show that $\lim\limits_{x\to 0}\,f(x)=0.$

We are interested in the behaviour of $f(x)$ when $x$ is close (but not equal) to $0$.  Note that $f(-x)=-f(x)$. So we will be finished if we can show that $f(x)$ approaches $0$ as $x$ approaches $0$ through positive values.

Let $x$ be small positive. Draw $\triangle OPQ$ as follows.  The base of the triangle is $OP$, and has length $1$.  The triangle is right-angled at $P$.  Finally, $Q$ is such that $\angle QOP =x$.  

Draw the circular sector with centre $O$, radius $1$, and going from $P$ to a point on $OQ$.  So the sector has angle $x$.   

Note that the circular sector is contained in $\triangle OPQ$.  The circular sector has area $(1/2)x$, and $\triangle OPQ$ has area $(1/2)\tan x$.  Thus the geometry gives us the inequality
$$\frac{x}{2}&lt;\frac{\tan x}{2}.$$
Since $x&gt;\sin x$, we get the estimates
$$0&lt;x-\sin x&lt; \tan x-\sin x.$$
The right-hand side only involves trigonometric functions, so is easier to deal with than $x-\sin x$:
$$\tan x-\sin x=\sin x\left(\frac{1-\cos x}{\cos x}\right)=\sin x\left(\frac{1-\cos^2 x}{\cos x(1+\cos x)}\right)=\frac{\sin^3 x}{\cos x(1+\cos x)}.$$
We conclude that 
$$0 &lt;\frac{x-\sin x}{x\sin x}&lt;\frac{\sin^2 x}{x\cos x(1+\cos x)}.$$
Since $\sin x&lt;x$, we find that 
$$0 &lt;\frac{x-\sin x}{x\sin x}&lt;\frac{\sin x}{\cos x(1+\cos x)},$$
and it is clear that $\dfrac{\sin x}{\cos x(1+\cos x)}$ approaches $0$ as $x$ approaches $0$ through positive values.

Comment: In this problem, there is no virtue in avoiding the calculus. The Taylor expansion is the natural  approach.
I did it that way:$$\lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\sin x}\right) =\lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) $$
because $\lim_{x\to0} \frac{\sin x}{x} = 1$ then
$$ \lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) \\= \lim_{x\to0} \left(\frac{\frac{\sin x}{x}* x - x}{\frac{\sin x}{x}*x}\right)\\= \lim_{x\to0} \left(\frac{x(\frac{\sin x}{x}* 1 - 1)}{\frac{\sin x}{x}*x}\right) \\=\lim_{x\to0} \left(\frac{\frac{\sin x}{x} - 1}{\frac{\sin x}{x}}\right) =[\frac{0}{1}] = 0 $$