How to evaluate $\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$? $$\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$$ 
without L'Hôpital.
I've tried using equivalences with ${(\sin(2x)-2\sin(x))^4}$ and arrived at $-x^{12}$ but I don't know how to handle ${(3+\cos(2x)-4\cos(x))^3}$. Using $\cos(2x)=\cos^2(x)-\sin^2(x)$ hasn't helped, so any hint?

Hint: Note that
$$ 3+\cos(2x)-4\cos(x) = 3 + 2\cos^2(x) - 1 - 4\cos(x) = 2(\cos(x)-1)^2, $$
and that
$$ \sin(2x) - 2\sin(x) = 2\sin(x)\cos(x)-2\sin(x) = 2\sin(x)(\cos(x)-1). $$
