$\frac{1}{\sin x}-\frac{1}{x}$ 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.
 A: 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 
$$ \lim_{x\to 0}\frac{x-\sin x}{x\sin x}=\lim_{x\to 0}\frac{1-\cos x}{\sin x+x\cos x}=\lim_{x\to 0}\frac{-\sin x}{2\cos x-x\sin x}=0.$$
A: 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.
A: You can also use L'Hopital. The limit as $x\to 0$ of 
$$\frac1{\sin x}-\frac 1x=\frac{x-\sin x}{x\sin x}$$
is of the type $0/0$. Hence, by L'Hopital it is equal to the limit as $x\to 0$ of
$$\frac{1-\cos(x)}{\sin(x)+x\cos(x)}.$$
Still $0/0$. Applying L'Hopital again we get
$$\frac{\sin(x)}{2\cos(x)-x\sin(x)}\longrightarrow\frac{0}{2}=0.$$
A: There is a cancellation of singularities. $\frac{1}{x}$ has a simple pole at $x=0$ and $\frac{1}{\sin x}$ has simple poles at $x\in\pi\mathbb{Z}$. By computing residues it follows that
$$ \frac{1}{\sin x}-\frac{1}{x}=2x\sum_{n\geq 1}\frac{(-1)^{n}}{x^2-\pi^2 n^2} $$
is given by a uniformly convergent series of continuous functions over $[-\pi/2,\pi/2]$.
A: Recall that $\lim_{x\to 0} \frac{\sin x}{x}=1$ so 
$\forall\varepsilon>0\,\exists\delta>0$ such that $\left|\frac{\sin x} x - 1\right|<\varepsilon$ for $x\in(-\delta,\delta)$ therefore
$$1-\varepsilon < \frac{\sin x} x < 1+\varepsilon$$
and
$$x(1-\varepsilon)<\sin x<x(1+\varepsilon) \text{ for } x\in(0,\delta)$$
Therefore for $x\in(0,\delta)$ your initial expression becomes
$$\frac1{x(1+\varepsilon)}-\frac1{x}<\frac1{\sin x}-\frac1{x}<\frac1{x(1-\varepsilon)}-\frac1x$$
reworking a little bit it gives
$$-\frac{\varepsilon}{x(1+\varepsilon)}<\frac1 {\sin x}-\frac 1 x <\frac{\varepsilon}{x(1-\varepsilon)}$$
A: As $\left({\pi\over2}\right)^2<6$ the theorem on alternating series gives us
$$\left|{1\over\sin x}-{1\over x}\right|=\left|{\sin x-x\over x\sin x}\right|\leq{x^3/6\over x\cdot{2\over\pi} x}={\pi x\over12}\leq{10\over24}\ \qquad\left(0<x\leq{\pi\over2}\right)\ .$$
