Finding the limit by using Maclaurin series $$ \lim_{x \to 0}\frac {x\cdot \sin x}{1-\cos(2x)} $$
I'm supposed to find the limit by using the terms, and figuring out $ x\times \sin x$ is easy enough, just multiply $\sin x$ with $x$, but when it's divided by $1-\cos(2x)$, I'm really lost. I can and have calculated what the first 5 terms of $1-\cos(2x)$ is, but I have no idea what I do with those two to get one series, so I can get the limit. If WolframAlpha isn't lying to me, the limit should be $\frac{1}{2}$.
From all my lecturer has written, I can't see anywhere that it is mentioned how I divide power series of any kind in the notes he wrote. Now, I'm studying over the internet, so I just follow the lectures online, and don't have any proper help around me, so hence, I'm asking here, for the second time in 24hrs.
So where do I go from having the first 5 terms of the Maclaurin series of $ x\times \sin x$ and $1-\cos(2x)$?

hint
$$\sin(x)=x+x\epsilon(x)$$
$$1-\cos(2x)=2\sin^2(x)$$
$$\frac{x\sin(x)}{1-\cos(2x)}=\frac{x}{2\sin(x)}$$
$$=\frac 12\frac{1}{\frac{\sin(x)}{x}}$$
$$=\frac 12\frac{1}{1+\epsilon(x)}$$
the limit is $ 0.5$.
