<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] We have that $$ \lim_{x \to 0} \frac{\sinh(x)}{\cosh(x)} - \frac{1}{x} = \lim_{x \to 0} \frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}$$ $$= \lim_{x \to 0} \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)} = \lim_{x \to 0} \frac{(\frac{x^3}{2!} + \frac{x^5}{4!} + \cdots) - (\frac{x^3}{3!} + \cdots) }{ ({x^2} + \frac{x^4}{3!} + \cdots) }$$ $$ = \lim_{x \to 0} \frac{(\frac{x}{2!}+\frac{x^3}{4!}+\cdots) - (\frac{x}{3!} + \cdots) }{(1 + \frac{x^2}{3!} + \cdots)} = 0 $$
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