<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] By the definition of hyperbolic functions and $e^x=1+x+\frac{x^2}2+o(x^2)$ we have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}=\frac{e^{2x}+1}{e^{2x}-1} - \frac{1}{x}=\\=\frac{2+2x+2x^2+o(x^2)}{2x+2x^2+o(x^2)} - \frac{1}{x}=\frac{2x+2x^2-2x-2x^2+o(x^2)}{2x^2+o(x^2)}=\frac{o(1)}{2+o(1)} \to 0$$
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