Let f: $R\to R$ be a function such that $\left|f\left(x\right)\right|\le x^2,$ for all $x\in R$ Then, at $x=0,$ f is:

$Let \left|f\left(x\right)\right|\le x^2,\forall x\in R$ 
Now, at  $x=0,\left|f\left(0\right)\right|\le 0$
$\Rightarrow f\left(0\right)=0$ 
$\therefore f\text{'}\left(0\right)=\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h-0}=\underset{h\to 0}{\lim }\frac{f\left(h\right)}{h}$ 
Now,  $\left|\frac{f\left(h\right)}{h}\right|\le \left|h\right|(\because \left|f\left(x\right)\right|\le x^2)$ 
$\Rightarrow -\left|h\right|\le \frac{f\left(h\right)}{h}\le \left|h\right|$ 
$\Rightarrow \underset{h\to 0}{\lim }\frac{f\left(h\right)}{h}\to 0$ 
(using sandwhich Theorem)  $\therefore$  from (i) and (ii), wer get  $f\text{'}\left(0\right)=0,$ 
i.e. - $f\left(x\right)$ is differentiable, at $x=0$ Since, differentiability $\Rightarrow$Continutity 
$\therefore \left|f\left(x\right)\right|\le x^2,$ for all  $x\in R$ is continuous as well as differentiable at  $x=0$