How to evaluate this limit using Taylor expansions? 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.

Using basic limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right)  &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\cos x-1}{x^{2}}\right)  &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\left( \frac{x}{e^{x}-1}\right)  &=&1
\end{eqnarray*}
which can be evaluated using L'HR, it follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{(\sin x-x)(\cos x-1)}{x(e^{x}-1)^{4}}
&=&\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{%
\cos x-1}{x^{2}}\right) \left( \frac{x}{e^{x}-1}\right) ^{4} \\
&=&\left( \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\right) \left(
\lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}}\right) \left( \lim_{x\rightarrow
0}\frac{x}{e^{x}-1}\right) ^{4} \\
&=&\left( -\frac{1}{6}\right) \left( -\frac{1}{2}\right) \left( 1\right) ^{4}
\\
&=&\frac{1}{12}.
\end{eqnarray*}
