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.
 A: Just in case you wanted something to check your answer off of:
\begin{align}
\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4} &= \lim_{x\to 0} \frac{\left(\frac{-x^3}{3!}+\frac{x^5}{5!} - \cdots\right)\left(\frac{-x^2}{2!} + \frac{x^4}{4!}- \cdots\right)}{x\left(x+\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right)^4}\\
&= \lim_{x\to 0} \frac{x^3\left(\frac{-1}{3!}+\frac{x^2}{5!} - \cdots\right)x^2\left(\frac{-1}{2!} + \frac{x^2}{4!}- \cdots\right)}{x^5\left(1+\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)^4}\\
&= \lim_{x\to 0} \frac{\left(\frac{-1}{3!}+\frac{x^2}{5!} - \cdots\right)\left(\frac{-1}{2!} + \frac{x^2}{4!}- \cdots\right)}{\left(1+\frac{x}{2!} + \frac{x^2}{3!} + \cdots\right)^4}\\
&= \left(\frac{-1}{3!}\right)\left(\frac{-1}{2!}\right)\\
&= \frac{1}{12}
\end{align}
A: Since we don't need Taylor Series at all, I thought it might be instructive to present a way forward that avoids series expansions.  
Note that we can write
$$\begin{align}
\frac{(\sin(x)-x)(\cos(x)-1)}{x(e^x-1)^4}&=-\frac{\frac{\sin(x)-x}{x^3}\,\frac{1-\cos(x)}{x^2}}{\left(\frac{e^x-1}{x}\right)^4}\\\\
&-\frac{\frac{\sin(x)-x}{x^3}\,\frac{2\sin^2(x/2)}{x^2}}{\left(\frac{e^x-1}{x}\right)^4}
\end{align}$$
Now, it is easy to show using the inequalities from elementary geometry
$$|x\cos(x)|\le |\sin(x)|\le |x|$$
for $|x|\le \pi/2$, that 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{2\sin^2(x/2)}{x^2}=\frac12}$$  
In addition, I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality that 
$$1+x\le e^x\le \frac{1}{1-x}$$
for $x<1$.  Then, it is easy to see that 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)^4=1}$$
We are left only to find the limit
$$\begin{align}
\lim_{x\to 0}\frac{\sin(x)-x}{x^3}&=\lim_{x\to 0}\frac{\cos(x)-1}{3x^2}\\\\
&=-\frac16
\end{align}$$
using L'Hospital's Rule once followed by using the aforementioned limit $\lim_{x\to 0}\frac{2\sin^2(x/2)}{x^2}=\frac12 $.
Putting it all together, we find the limit of interest is 
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{(\sin(x)-x)(\cos(x)-1)}{x(e^x-1)^4}=\frac{1}{12}}$$
A: Turns out to just be a foolish mistake. When I factored $x$ out of the denominator I should have raised it to 4. That would be an $x^5$ on the bottom:
$\lim_{x \to 0} \dfrac{x^3(-\frac{1}{3!}+\frac{x^2}{5!}...)x^2(-\frac{9}{2!}+\frac{81x^2}{4!}...)}{x^5({1}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}...)^4}$
Now I can simply cancel out the $x^5$s and then substitute remaining $x$s for zeroes.
$\lim_{x \to 0} \dfrac{(-\frac{1}{3!})(-\frac{9}{2!})}{({1})^4} = \dfrac{3}{4} $
A: 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*}
