How can this trigonometric inequality related to a limit be proved? I want to prove that $\;\;\displaystyle \left|\frac{\sin x-x}{x^{2}}\right|\leq\frac{4(\pi/2-1)}{\pi^{2}}\;\;$
for all $x$ such that $x\in\left[0,\pi/2\right]$.
If you look at the graph of the expression on the left, it is clearly (appearing to be) monotonically increasing, so the maximum value of the left hand side is the output of the expression when $x=\pi/2$.
Would like a proof that does not involve calculus since this inequality is being used to prove $\displaystyle\lim_{x\rightarrow0}\frac{\sin x}{x}=1$.
 A: You may already know that $\sin x < x< \tan x$ for $0<x<\frac\pi 2$.
Then for such $x$ we have
$$0< \frac{x-\sin x}{x^2}< \frac{\tan x-\sin x}{x^2}=\frac{(1-\cos x)\tan x}{x^2}=\frac{\sin^2x\tan x}{x^2(1+\cos x)}< \tan x.$$
This is not the bound you were asking for, but it is weaker only for big $x$, hence is  more than   enough to show $\lim_{x\to 0}\frac{\sin x}{x}=1$. (In fact, using $\cos x\to 1$, we obtain $\sin x=x+O(x^3)$)
A: Actually we can expect this function 
$$f(x)=\dfrac{x -\sin x }{x^2}$$
as an increasing function.
$f'(x) = \dfrac{2\sin x-x\cos x-x}{x^3}$
we just have to prove 
$g(x) = 2\sin x-x\cos x-x\ge 0$ , for $x\in[0,\pi/2]$
since $g(0) = 0$ , we just prove $g(x)$ is also increasing on $[0,\pi/2]$.
$g'(x) = \cos x+x\sin x -1$, with $g'(0) = 0$
and $g''(x) = x\cos x\ge 0$ on the interval $[0,\pi/2]$, thus $g'(x)$ is non-decreasing, thus $g'(x)\ge 0$, which means $g(x)$ is non-decreasing, thus $g(x)\ge 0.$
Therefore,
$f'(x)\ge 0$, $f$ is non-decreasing, thus $f(\dfrac{\pi}{2})$ will be the maximum.
