$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ as $x$ goes to $0$ 
Calculate $\displaystyle \lim_{x \to 0}\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$

I tried L'hopital, but the denominator gets more and more complicated. 
How does one calculate this limit? 

Hint:$\cos 2x^2=1-2\sin^2 x^2$, and  replacing this in the numerator gives us an expression equivalent to $\frac{-2\sin x^2}{x^2}$. Look familiar?
