$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ 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? 

Recall that $\cos(2t) = 1-2\sin^2(t)$. Hence, we have
$$\cos(2x^2)-1 = -2\sin^2(x^2)$$
Hence, we have
$$\lim_{x \to 0} \dfrac{\cos(2x^2)-1}{x^2\sin(x^2)} = \lim_{x \to 0} \dfrac{-2\sin^2(x^2)}{x^2\sin(x^2)} = -2 \lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = -2$$
