So we're supposed to calculate the limit of $\frac{x\sin(x)}{1-\cos(x)}$ as $x\to 0$ using power series.
Using the series expansions for $\sin(x)$/$\cos(x)$ and multiplying in the $x$ I get
$$\frac{\sum_0^\infty\frac{x^{2n+2}(-1)^{n}}{(2n+1)!}} {1-\sum_0^\infty \frac{x^{2n}(-1)^{n}}{(2n)!}} =\frac{\sum_1^\infty\frac{x^{2n}(-1)^{n-1}}{(2n-1)!}} {\sum_1^\infty \frac{x^{2n}(-1)^{n+1}}{(2n)!}}$$
Using l'Hôpital on the initial expression twice, it's easy to see that the limit is 2.
So I'd expect to find only an extra factor of $2$, not $2n$ for every term in the denominator, like this: $$\frac{\sum_1^\infty\frac{x^{2n}(-1)^{n-1}}{(2n-1)!}} {\sum_1^\infty \frac{x^{2n}(-1)^{n+1}}{2(2n-1)!}}$$ However, WA tells me that the RHS above is still correct. Why does the $n$ not disturb the result?
I'm sorry, I might have posed the question too early:
When separating the numerator, we get:
$$\frac{x^2}{\sum_1^\infty \frac{x^{2n}(-1)^{n+1}}{(2n)!}}-\frac{\frac{x^4}{6}}{\sum_1^\infty \frac{x^{2n}(-1)^{n+1}}{(2n)!}}+\dots$$
So if we then cancel the n-th fraction by $x^{2n}$, only the first fraction remains with no elements in the denominator which diverge towards $\infty$ as $x$ approaches $0$:
$$\frac{x^2}{\frac{x^2}{2}-\frac{x^4}{4\cdot 6}+\dots}-\frac{\frac{x^4}{6}}{\frac{x^2}{2}-\frac{x^4}{4\cdot 6}+\dots}+\dots$$ $$\implies \frac{1}{\frac{1}{2}-\frac{x^2}{4\cdot 6}+\dots}-\frac{\frac{1}{6}}{\frac{1}{2x^2}-\frac{1}{4\cdot 6}+\dots}+\dots$$
So that only the first term "survives", and there the ratio is $2$. The thing I don't know is how to write this argument in closed form.