Question Number 179503 by cortano1 last updated on 30/Oct/22
$$\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\frac{\pi}{\mathrm{4}}}\:−\frac{\mathrm{2}}{\mathrm{x}−\mathrm{1}}\right)\:=? \\ $$
Commented by CElcedricjunior last updated on 30/Oct/22
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\boldsymbol{\mathrm{x}}−\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{arctan}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\pi}/\mathrm{2}}{\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{arctanx}}−\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)}\right)=\frac{\mathrm{0}}{\mathrm{0}}=\boldsymbol{\mathrm{FI}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{apply}}\:\boldsymbol{{hospital}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{arctan}}\left(\boldsymbol{{x}}\right)−\frac{\boldsymbol{\pi}}{\mathrm{2}}+\frac{\boldsymbol{\mathrm{x}}−\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\frac{\mathrm{4}\boldsymbol{\mathrm{x}}}{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }+\frac{\mathrm{1}+\mathrm{2}\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{1} \\ $$