<s>[INST] Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}= \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)}$$
but I don't know how to reduce this further.
 [/INST] Near $0$, you have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x} = \frac{1 + \frac{x^2}{2} + o(x^3)}{x + \frac{x^3}{6} + o(x^4)} - \frac{1}{x} = \frac{1}{x}\left[\left( 1 + \frac{x^2}{2} + o(x^3) \right)\left( 1 + \frac{x^2}{6} + o(x^3) \right)^{-1}-1\right]$$ $$ =\frac{1}{x}\left[\left( 1 + \frac{x^2}{2} + o(x^3) \right)\left( 1 - \frac{x^2}{6} + o(x^2) \right)-1\right] =\frac{1}{x}\left( \frac{x^2}{3} + o(x^2) \right) = \frac{x}{3} + o(x) $$
So the limit is $0$.
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