$\frac{1}{\sin x}-\frac{1}{x}$ 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.

The problem is only near $x=0$. We have
$$\frac1{\sin x}-\frac 1x=\frac{x-\sin x}{x\sin x}.$$
As $\sin x = x+O(x^3)$, this is $\frac {O(x^3)}{x^2+O(x^4)} =O(x)$.
More explicitly, we use l'Hôpital to see that 
$$ \lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{1-\cos x}{\sin x+x\cos x}=\lim_{x\to 0}\frac{-\sin x}{2\cos x-x\sin x}=0.$$
