How to evaluate $\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$?

Question

$$\lim_{x\to 0} \frac {(\sin(2x)-2\sin(x))^4}{(3+\cos(2x)-4\cos(x))^3}$$

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?

Answer 1

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 $$ \sin(2x) - 2\sin(x) = 2\sin(x)\cos(x)-2\sin(x) = 2\sin(x)(\cos(x)-1). $$

Answer 2

Hint: Your quotient can be simplified to $$8\cos\left(\frac{x}{2}\right)^4$$

Answer 3

One can evaluate all such lims using series expansions. $$ \sin(x) = x - \frac{x^3}{6} + o(x^4)\\ \cos(x) = 1 - \frac{x^2}{2} + o(x^4) $$ https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions

Just substitute functions with their expansions. Then just find lim of expression~polynomial/polynomial. Add more $x^n$ terms, if first 2 is not enough.

L'Hôpital's rule is another form of this more general approach.