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.
 A: \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}
A: We have that $$ \lim_{x \to 0} \frac{\sinh(x)}{\cosh(x)} - \frac{1}{x} = \lim_{x \to 0} \frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}$$ $$= \lim_{x \to 0} \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)} = \lim_{x \to 0} \frac{(\frac{x^3}{2!} + \frac{x^5}{4!} + \cdots) - (\frac{x^3}{3!} + \cdots) }{ ({x^2} + \frac{x^4}{3!} + \cdots) }$$ $$ = \lim_{x \to 0} \frac{(\frac{x}{2!}+\frac{x^3}{4!}+\cdots) - (\frac{x}{3!} + \cdots) }{(1 + \frac{x^2}{3!} + \cdots)} = 0 $$
A: Reduce to the same denominator and use L'Hospital (I do not recommend it so often…):
$$f(x)=\frac{\cosh x}{\sinh x} - \frac{1}{x}=\frac{x\cosh x-\sinh x}{x\sinh x}, $$
therefore
\begin{align}\lim_{x\to 0}f(x)&=\lim_{x\to 0}\frac{\cosh x+x\sinh x-\cosh x}{\sinh x+x\cosh x}
=\lim_{x\to 0}\frac{x\sinh x}{\sinh x+x\cosh x}\\[1ex]
&=\lim_{x\to 0}\frac{\sinh x\;\}\scriptstyle\to 0}{\cfrac{\sinh x}{\underbrace{\quad  x\quad}_{\to 1}}+\underbrace{\cosh x}_{\to 1}}.
\end{align}
A: 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 $0$.
A: By the definition of hyperbolic functions and $e^x=1+x+\frac{x^2}2+o(x^2)$ we have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}=\frac{e^{2x}+1}{e^{2x}-1} - \frac{1}{x}=\\=\frac{2+2x+2x^2+o(x^2)}{2x+2x^2+o(x^2)} - \frac{1}{x}=\frac{2x+2x^2-2x-2x^2+o(x^2)}{2x^2+o(x^2)}=\frac{o(1)}{2+o(1)} \to 0$$
