Question Number 214256 by Einstein2006 last updated on 02/Dec/24
$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\:\rightarrow\:\mathrm{1}} \propto.\boldsymbol{{arctan}}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\boldsymbol{{x}}}\:−\:\mathrm{1}\right) \\ $$$$\bullet\:\boldsymbol{{Calculons}}\:\boldsymbol{{la}}\:\boldsymbol{{limite}}\:\boldsymbol{{a}}\:\boldsymbol{{l}}'\boldsymbol{{intrieur}}: \\ $$$$\frac{\mathrm{2}}{\mathrm{1}\:+\:\boldsymbol{{x}}}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\:\rightarrow\:\mathrm{1}} {arctan}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\:{x}}\:−\:\mathrm{1}\right)=\:\boldsymbol{{arctan}}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{1}} \propto.{arctan}\left(\frac{\mathrm{2}}{\mathrm{1}+\:{x}}\:−\:\mathrm{1}\right)\:=\propto.\mathrm{0}\: \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{1}} \propto.\boldsymbol{{arctan}}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\:\boldsymbol{{x}}}\:−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$