$\frac{1}{\sin x}-\frac{1}{x}$ bounded on $[0,\pi/2]$. Why is $$\frac{1}{\sin x}-\frac{1}{x}$$ bounded when $x\in [0,\pi/2]$. I've come across this fact in Fourier series, but I can't figure out a why this is true. I would appreciate any help.

This function is continuous on the half-open interval $(0,\,\pi/2].$  At $x=0$ it can be extended so as to be continuous by letting its value by its limit as $x\to0$ if that exists. Then one has a continuous function on a closed bounded interval, and such functions are bounded. And if that extended function is bounded, then so is the function you started with before doing the extension.
So the question is whether the following exists (meaning it's a finite number and not $+\infty$ or $-\infty$):
\begin{align}
\lim_{x\to0} \left( \frac 1 {\sin x} - \frac 1 x \right) = \lim_{x\to0} \frac{x-\sin x}{x\sin x}.
\end{align}
L'Hopital's rule handles that.
