Evaluate $ \lim\limits_{x \to 0} \frac{1}{\sin^3x}\int_0^x{\sin(t^2) } dt$ 
$$ \lim_{x \to 0} \frac{1}{\sin^3x}\int_0^x{\sin(t^2) dt}$$  

This is what I've tried: 
Let $F(x) = \displaystyle\int_0^x{\sin(t^2) dt}$, and let $f(x) = {\sin(t^2)}$.
Then $F'(x) = f(x) \Rightarrow F'(0) = f(0) = \sin 0 = 0$. 
Where do I go from here?
Thanks for help.

Using L'Hospital's rule once, we have
$$
\lim_{x\rightarrow 0}\frac{\sin x^2}{3\sin^2x \cos x}.
$$
We examine the limit
$$
\lim_{x\rightarrow 0} \frac{\sin x^2}{\sin^2x} = \lim_{x\rightarrow 0} \frac{2x\cos x^2}{2\sin x \cos x} =\lim_{x\rightarrow 0} \frac{x\cos x}{\sin x} = 1
$$
using the fact that $\lim_{x\rightarrow 0} \frac{x}{\sin x} = 1$. Thus,
$$
\lim_{x\rightarrow 0}\frac{\sin x^2}{3\sin^2x \cos x} = \lim_{x\rightarrow 0} \frac{1}{3\cos x} = \frac{1}{3}.
$$
