Question Number 130401 by john_santu last updated on 25/Jan/21
$${Where}\:{is}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{ln}\:{x}+\mathrm{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{2}} −\mathrm{1}}\:{continous}? \\ $$
Answered by mathmax by abdo last updated on 25/Jan/21
![f(x)=((lnx+arctanx)/(x^2 −1)) f is defined on ]0,1[∪]1,+∞[ =D f is contnue on D is f continue on 1? lim_(x→1^+ ) f(x)=((π/4)/0^+ )=+∞ and lim_(x→1^− ) f(x)=−∞ ⇒f is not onx_0 =1](https://www.tinkutara.com/question/Q130430.png)
$$\left.\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{lnx}+\mathrm{arctanx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\:\mathrm{f}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{on}\:\right]\mathrm{0},\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\:=\mathrm{D}\right. \\ $$$$\mathrm{f}\:\mathrm{is}\:\mathrm{contnue}\:\mathrm{on}\:\mathrm{D}\:\mathrm{is}\:\mathrm{f}\:\mathrm{continue}\:\mathrm{on}\:\mathrm{1}? \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\frac{\pi}{\mathrm{4}}}{\mathrm{0}^{+} }=+\infty\:\:\mathrm{and}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{−} } \:\:\mathrm{f}\left(\mathrm{x}\right)=−\infty\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{onx}_{\mathrm{0}} =\mathrm{1} \\ $$