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Question Number 38206 by prof Abdo imad last updated on 22/Jun/18

$$calculate{lim}_{x\to 0}\frac{xcoth\left(x\right)-1}{x^2}$$

Commented by math khazana by abdo last updated on 25/Jun/18

$$wehaveprovedthat$$$$coth\left(x\right)-\frac{1}{x}=\sum _{n=1}^{\mathrm{\infty }}\frac{2x}{x^2+n^2{\pi }^2}(x\ne o)\Rightarrow$$$$\frac{xcoth\left(x\right)-1}{x^2}=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{x^2+n^2{\pi }^2}\Rightarrow$$$${lim}_{x\to 0}\frac{xcoth\left(x\right)-1}{x^2}=\frac{2}{{\pi }^2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$$$$=\frac{2}{{\pi }^2}\frac{{\pi }^2}{6}=\frac{1}{3}.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

$$\underset{x\to 0}{\lim }\frac{x\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)-1}{x^2}$$$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$$$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$$$$x\left(\frac{e^x+e^{-x}}{e^x-e^{-x}}\right)=x\left(\frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+...}{x+\frac{x^3}{3!}+\frac{x^5}{5!}+...}\right)=\frac{1+\frac{x^2}{2!}+...}{1+\frac{x^2}{3!}+..}$$$$\underset{x\to 0}{\lim }\frac{\frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+..}{1+\frac{x^2}{3!}+\frac{x^4}{5!}+..}-1}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{x^2(\frac{1}{2!}-\frac{1}{3!})+x^{2}f\left(x\right)}{x^2}$$$$whenx\to 0f\left(x\right)\to 0$$$$=\frac{1}{2}-\frac{1}{6}=\frac{3-1}{6}=\frac{1}{3}\mathcal{A}NS$$$$$$

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