<s>[INST] 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.
 [/INST] 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}
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