$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ 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? 
 A: Recall that $\cos(2t) = 1-2\sin^2(t)$. Hence, we have
$$\cos(2x^2)-1 = -2\sin^2(x^2)$$
Hence, we have
$$\lim_{x \to 0} \dfrac{\cos(2x^2)-1}{x^2\sin(x^2)} = \lim_{x \to 0} \dfrac{-2\sin^2(x^2)}{x^2\sin(x^2)} = -2 \lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = -2$$
A: 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?
A: Notice, you can find this limit by using 'Hôpital's rule (do it twice):
$$\lim_{x\to0}\frac{\cos(2x^2)-1}{x^2\sin(x^2)}=\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(\cos(2x^2)-1\right)}{\frac{\text{d}}{\text{d}x}\left(x^2\sin(x^2)\right)}=$$
$$\lim_{x\to0}\frac{-4x\sin(2x^2)}{2x^3\cos(x^2)+2x\sin(x^2)}=-2\left[\lim_{x\to0}\frac{\sin(2x^2)}{x^2\cos(x^2)+\sin(x^2)}\right]=$$
$$-2\left[\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(\sin(2x^2)\right)}{\frac{\text{d}}{\text{d}x}\left(x^2\cos(x^2)+\sin(x^2)\right)}\right]=-2\left[\lim_{x\to0}\frac{4x\cos(2x^2)}{4x\cos(x^2)-2x^3\sin(x^2)}\right]=$$
$$-2\left[\lim_{x\to0}-\frac{2\cos(2x^2)}{x^2\sin(x^2)-2\cos(x^2)}\right]=-2\left[-\frac{2\cos(2\cdot0^2)}{0^2\sin(0^2)-2\cos(0^2)}\right]=$$
$$-2\left[-\frac{2\cdot1}{0-2\cdot1}\right]=-2\left[-\frac{2}{-2}\right]=-2\left[--\frac{2}{2}\right]=-2\left[\frac{2}{2}\right]=-2$$
