<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] \begin{align}
& \frac{\left(x\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\right)-\left(x + \frac{x^3}{3!} + \cdots\right)}{x\left(x + \frac{x^3}{3!}+\cdots\right)} \\ {} \\
= {} & \frac{\left(x + \frac{x^3}{2!} + \frac{x^5}{4!}+ \cdots\right) -\left( x + \frac{x^3}{3!} \right)}{x^2 + \frac{x^4}{3!}+ \cdots} \\ {} \\
= {} & \frac{\left(\frac{x^3}{2!} + \frac{x^5}{4!}+ \cdots\right) -\left(\frac{x^3}{3!} + \cdots \right)}{x^2 + \frac{x^4}{3!}+ \cdots} \\ {} \\
= {} & \frac{\left(\frac{x}{2!} + \frac{x^3}{4!}+ \cdots\right) -\left(\frac x {3!}+ \cdots \right)}{1 + \frac{x^2}{3!}+ \cdots} \to \frac 0 1 
\end{align}
</s>