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 $ ? 


Given $\varepsilon >0$ there exist $\delta_{\varepsilon}>0$ such that 
$$-\delta_{\varepsilon}<x<\delta_{\varepsilon}\qquad \implies \qquad\left|\frac{f(x)}{x^2}-L\right|<\varepsilon$$
So $$|f(x)-Lx^2|<\varepsilon x^2$$
Now, take $\delta=\min(\delta_{\varepsilon},1)$, then
$$-\delta<x<\delta\qquad \implies \qquad\left|f(x)-Lx^2\right|<\varepsilon\tag{1}$$
So 
\begin{align}
\lim_{x\to0}f(x)&=\lim_{x\to0}Lx^2=0\qquad\text{from }(1)
\end{align}
Since $f$ is continuous at $0$ we get
$$f(0)=\lim_{x\to0}f(x)=0$$
