<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] 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} $
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