If $f(x)$ is continuous at $ x=0$ Given that  $f(x)$ is continuous at $ x=0$, and the limit : $$\lim \limits_{x \to 0} \frac{f(x)}{x^2} = L$$
then:
$$\implies f(0) = 0 $$
and 
 $$ \implies f(x) \text{ is differentiable at }x=0 $$

My question is: why $f(0) = 0 $ ? 


If $f(0)\neq 0$, then $\frac{f(x)}{x^2}$ diverges, since $$\lim_{x\to 0}\frac{c}{x^2}=\pm\infty$$
