Question Number 197008 by tri26112004 last updated on 06/Sep/23
Answered by AST last updated on 06/Sep/23
$$\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={Z} \\ $$$${Along}\:{y}={x},{Z}=\mathrm{0};{Along}\:{x}=\mathrm{0},{Z}=−\mathrm{1} \\ $$$$\Rightarrow{lim}\:{does}\:{not}\:{exist} \\ $$
Answered by MM42 last updated on 07/Sep/23
$$\left.{a}\right)\:{if}\:\:{x}={y}\Rightarrow{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} \:\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$$\:{if}\:\:{y}=\mathrm{2}{x}\Rightarrow{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} \:\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}{x}^{\mathrm{2}} }=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow\:{lim}\:\:“\:{not}\:{exist}\:''\:\checkmark \\ $$$$\left.{b}\right)\:\:{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{2}\right)} \:\frac{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} \right)\left(\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\ $$