I have just learnt the hyperbolic function and I meet a problem goes like follows: $$\lim_{x\rightarrow 0}k^2\operatorname{csch}^2(kx)-\frac{1}{4}\operatorname{csch}^2\left(\frac{x}{2}\right)$$ I have tried refomulating it as a form of fraction and applying the L'Hopital's Law but it seems that it just leads the problem to a much more messy land, because multiple differentiations is required. I wonder if there exists any easy way to solve it?
Any comments will be appreciated.
On
Taylor expansion is very fine, if we want to proceed by l'Hopital we have
$$\lim_{x\rightarrow 0} \frac1{\sinh^2 x}-\frac1{x^2}=\lim_{x\rightarrow 0} \frac{x^2-\sinh^2 x}{x^2\sinh^2 x}\stackrel{H.R.}=\lim_{x\rightarrow 0}\frac{x-\sinh x\cosh x}{x\sinh^2 x+x^2\sinh x\cosh x}=$$
$$\stackrel{H.R.}=\lim_{x\rightarrow 0}\frac{1-\sinh^2 x-\cosh^2 x}{\sinh^2 x+4x\sinh x\cosh x+x^2\cosh^2 x+x^2\sinh^2 x}=$$
$$\stackrel{\text{dividing by }x^2}=\lim_{x\rightarrow 0}\frac{-2\frac{\sinh^2 x}{x^2}}{\frac{\sinh^2 x}{x^2}+4\frac{\sinh x}{x}\cosh x+\cosh^2 x+\sinh x}=\frac{-2}{1+4+1+0}=-\frac13$$
and then
$$ k^2\operatorname{csch}^2(kx)-\frac{1}{4}\operatorname{csch}^2\left(\frac{x}{2}\right)=\frac{k^2}{\sinh^2(kx)}-\frac{1}{4\sinh^2\left(\frac{x}{2}\right)}=$$
$$=\frac{k^2}{\sinh^2(kx)}-\frac{k^2}{(kx)^2}-\frac{1}{4\sinh^2\left(\frac{x}{2}\right)}+\frac{1}{4\left(\frac{x}{2}\right)^2}\to-\frac{k^2}{3}+\frac1{12}$$
Hint. We can reason with truncated Taylor series:
$$ \begin{array}{lll} & \sinh(x) & \displaystyle =x+\frac{x^3}{6}+\cdots \\[3pt] \implies & \mathrm{csch}(x) & \displaystyle =\frac{1}{x}\frac{1}{1+\frac{x^2}{6}+\cdots} \\[1pt] & & \displaystyle =\frac{1}{x}\Bigl(1-\frac{x^2}{6}+\cdots\Bigr) \\[1pt] & & \displaystyle=\frac{1}{x}-\frac{x}{6}+\cdots \\[4pt] \implies & \mathrm{csch}^2(x) & \displaystyle = \frac{1}{x^2}-\frac{1}{3}+\cdots \end{array} $$
The middle part follows from the geometric series formula $\frac{1}{1+u}=1-u+u^2-\cdots$
You can replace all the "$\cdots$"s with big-$\mathcal{O}$ notation if desired.