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Question Number 130401 by john_santu last updated on 25/Jan/21

$$Whereisthefunction$$$$f\left(x\right)=\frac{\ln x+{\tan }^{-1}x}{x^2-1}continous?$$

Answered by mathmax by abdo last updated on 25/Jan/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{lnx}+\mathrm{arctanx}}{{\mathrm{x}}^2-1}\mathrm{f}\mathrm{is}\mathrm{defined}\mathrm{on}]0,1[\cup ]1,+\mathrm{\infty }[=\mathrm{D}$$$$\mathrm{f}\mathrm{is}\mathrm{contnue}\mathrm{on}\mathrm{D}\mathrm{is}\mathrm{f}\mathrm{continue}\mathrm{on}1?$$$${\lim }_{\mathrm{x}\to 1^{+}}\mathrm{f}\left(\mathrm{x}\right)=\frac{\frac{\pi }{4}}{0^{+}}=+\mathrm{\infty }\mathrm{and}{\lim }_{\mathrm{x}\to 1^{-}}\mathrm{f}\left(\mathrm{x}\right)=-\mathrm{\infty }\Rightarrow \mathrm{f}\mathrm{is}\mathrm{not}{\mathrm{onx}}_0=1$$

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