How to show $ \Big\vert \frac{\sin(x)}{x} \Big\vert $ is bounded by $1$? This may be a silly question, but I cannot figure it out. I want to prove that
$ \Big\vert \frac{\sin(x)}{x} \Big\vert \leq 1 $ for $x\in[-1,0)\cup(0,1]$,
but I don't even know where to start.

Hint By the MVT you have
$$\frac{\sin(x)-\sin(0)}{x-0} =\cos(c)$$
for some $c$.
