Proving the function $\frac{1-\cos(x)}{x^2}$ is bounded. How can I prove that $0\leq{f(x)}<\frac{1}{2}$ for every $x\neq{0}$, when
$$f(x)=
\begin{cases}
\frac{1-\cos(x)}{x^2}, & x\neq{0} \\
\frac{1}{2}, & x=0
\end{cases}
$$
I know how to see that $0\leq{f(x)}$ but I dont know how to prove the other boundary.
 A: Note that for $x\neq 0$
$$1-\frac{x^2}2< \cos x \le 1$$
thus
$$0\le \frac{1-\cos x}{x^2}< \frac12$$
Observe that $\cos x>1-\frac{x^2}2$ can be easily shown by MVT.
Using mean value theorem to show that $\cos (x)>1-x^2/2$
A: $\cos(x) = 1 - x^2/2 + o_{x \to 0}(x^2)$
Hence $f(x) \to_{x \to 0} f(0) = 1/2$ 
Then you could derivate $f$ on $(0, 1/2)$ to see it is strictly decreasing.
A: Since $1-\cos x = 2 \sin ^2 (x/2)$, replacing $x=2t$, we only need to show 
$$\left|\frac{\sin t}{t}\right|<1\tag{1}$$
for all $t$. This is sort of classical. Here I only sketch an elementary proof.

Apparently, we only need to show (1) for $|t|<1$, which conveniently lies in $|t|<\pi/2$. Since $\sin t> 0$ for $0<t<\pi/2$ and $\sin(-t) = - \sin t$, we only need to show
$$\sin t < t\tag{2}$$
for $0<t<\pi/2$.
Now (2) is easy, given the derivative of $g(t) = \sin t - t$ is $g'(t) = \cos t - 1\le 0$. 
A: Recall that by construction
$$0\le\frac{\sin x}x\le1$$
For first-quadrant angles $x$. Then it follows that for first or second quadrant angles,
$$0\le\left(\frac{\sin\frac x2}{\frac x2}\right)^2\cos^2\frac x2\le\cos^2\frac x2$$
Equivalently,
$$0\le\frac{\sin^2x}{x^2}=\frac{(1+\cos x)(1-\cos x)}{x^2}\le\frac12(1+\cos x)$$
We can verify directly if $x=\frac{\pi}2$ or by division for other first or second quadrant angles that
$$0\le f(x)=\frac{1-\cos x}{x^2}\le\frac12$$
Because this is an even function of $x$ it's true for $x\in[-\pi,0)\cup(0,\pi]$ and if we define $f(0)=\frac12$, then it's true for $x=0$ as well. Also it's true for $|x|>\pi$ because then $x^2>4$ and $1-\cos x\le2$. And equality at the upper limit only happens if $\frac{\sin x}x=1$ which is only true in the limit as $x\rightarrow0$ or when $x=(2n+1)\pi$ but we can see by inspection that equality doesn't hold in that case either, so the upper limit is strict except when $x=0$ by the definition of $f(x)$.
