Question Number 72838 by Rio Michael last updated on 03/Nov/19
$${given}\:{that}\:\:\:{f}\left({x}\right)\:=\:\frac{\mid{x}\:−\mathrm{2}\mid}{\mathrm{1}−\mid{x}\mid} \\ $$$${check}\:{if}\:{f}\:{is}\:{continuous}\:{a}\:{x}\:=\:\mathrm{2} \\ $$$${hence}\:\:{write}\:{f}\left({x}\right)\:{as}\:{a}\:{pairwise}\:{function}\: \\ $$
Commented by mathmax by abdo last updated on 03/Nov/19
$${f}\left(\mathrm{2}\right)=\mathrm{0}\:\:\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{2}^{+} } \:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{2}^{+} } \:\:\frac{{x}−\mathrm{2}}{\mathrm{1}−\mathrm{2}}\:=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{2}^{−} } \:\:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{2}^{−} } \:\:\:\:\frac{−{x}+\mathrm{2}}{\mathrm{1}−\mathrm{2}}\:=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{2}^{−} } \:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{2}^{+} } \:{f}\left({x}\right)={f}\left(\mathrm{2}\right)\:\Rightarrow{f}\:{is}\:{continue}\:{at}\:{x}_{\mathrm{0}} =\mathrm{2} \\ $$