calculus - Solving $limlimits_{xto0} frac{x - sin(x)}{x^2}$ without L'Hospital's Rule.

How to solve $\lim\limits_{x\to 0} \frac{x - \sin(x)}{x^2}$ Without L'Hospital's Rule?
you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.

Answer

The given expression is odd; therefore it is enough to consider $x>0$. We then have

$$0<{x-\sin x\over x^2}<{\tan x -\sin x\over x^2}=\tan x\ {1-\cos x\over x^2}={\tan x\over2}\ \Bigl({\sin(x/2)\over x/2}\Bigr)^2\ ,$$
and right side obviously converges to $0$ when $x\to0+$.