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 $1-\cos(2x)$?

We have that $\cos 2x = 1 - \frac12 (2x)^2 + \frac1{24} (2x)^4 + \dots,$ so $1 - \cos 2x = 2 x^2 - \frac{2}{3} x^4 + \dots$ and $x \sin x = x^2 - \frac16 x^4 + \dots.$ Thus $\lim_{x \to 0} \frac{x \sin x}{1 - \cos 2x} = \frac{x^2 - \frac16 x^4 + \dots}{2 x^2 - \frac23 x^4 + \dots} = \lim_{x \to 0} \frac{1 - \frac16 x^2 + \dots}{2 - \frac23 x^2 + \dots} = \frac12$ simply by evaluating the limits of the numerator and denominator. Hope this helps. :)
