I am trying to evaluate this limit:

$\lim\_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$

I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their first term removed so the series starts at $n=1$ rather than at 0.

However, when I actually try to evaluate I get stuck at:

$\lim\_{x \to 0} \dfrac{(-\frac{1}{3!}+\frac{x^2}{5!}...)(-\frac{9x^2}{2!}+\frac{81x^4}{4!}...)}{(\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}...)^4}$

I don't know how to proceed from here.