<s>[INST] Evaluating the Limit with Series Expansion Problem
Evaluate the following using series expansion.
$$ \lim_{x \rightarrow 0} \frac{\cosh{x}-\cosh{2x}}{x \cdot \sinh{x} }$$

I really don't know what I should do. If I expand $\cosh$ and $\sinh$ normally, (they'd look something like $\sum_{n = 0}^{\infty}\frac{x^{2n}}{(2n)!}$) I get a fraction with two infinite sums with the limit $0$ on both top and bottom and if I expand it using Taylor, I have to use $x \neq 0$ and I get a sum with infinite summands.
I don't have anymore ideas and any help would be appreciated. Thank you in advance.
 [/INST] As other answers said, you are correct and the limit is easy to find using this way.
However, the same procedure can be used to see how is approached the limit. Just use one extra term in the expansions $$\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$  $$\cosh(2x)=1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right)$$ $$\sinh(x)=x+\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{\left(1+\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right) \right)-\left(1+2 x^2+\frac{2 x^4}{3}+O\left(x^6\right) \right) } { x^2+\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^7\right)}$$
$$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=\frac{-\frac{3 x^2}{2}-\frac{5 x^4}{8}+O\left(x^6\right) }{ x^2+\frac{x^4}{6}+\frac{x^6}{120}+O\left(x^7\right) }$$ Simplify by $x^2$ and use long division to finally get $$\frac{\cosh(x)-\cosh(2x)}{x\sinh(x)}=-\frac{3}{2}-\frac{3 x^2}{8}+O\left(x^4\right)$$
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