Calculate: $\lim_{x\to 0} \frac{f(x^2)-f(0)}{\sin^2(x)}$. Let $f(x)$ be a differentiable function. s.t. $f^\prime(0)=1$. 
calculate the limit: $$\lim_{x\to 0} \frac{f(x^2)-f(0)}{\sin^2(x)}.$$
SOLUTION ATTEMPT: I thought that because $f$ is differentiable its also continuous, then we can say: $\lim_{x\to 0} f(x^2)=f(0)$ then, $\lim_{x\to 0} f(x^2)-f(0)=0$ and also $\lim_{x\to 0} \sin^2(x)=0$, so using L'Hoptal's rule, we get that:
$\lim_{x\to 0 } \frac{f(x^2)-f(0)}{\sin^2(x)}= \lim_{x\to 0} \frac{f^\prime (x^2) \cdot 2x}{2\sin(x) \cdot \cos(x)}=\lim_{x\to 0} \frac{f^\prime (x^2) \cdot 2x}{2\sin(x) \cdot \cos(x)}$.
I reached right here and I guess I need to do another L'Hopital, is that the right direction? 

You are done there, since
$$
\lim_{x\to 0} \frac{f'(x^2)\cdot 2x}{2\sin x\cos x} = \lim_{x\to 0} \frac{2x}{2\sin x} \cdot \lim_{x\to 0}\frac{f'(x^2)}{\cos x} = 1\cdot 1 =1.$$
