Question Number 187611 by otchereabdullai last updated on 19/Feb/23
$$\:{Use}\:{polar}\:{coordinate}\:{to}\:{find} \\ $$$${lim}\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)\:\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$
Answered by mahdipoor last updated on 19/Feb/23
$${not}\:{exist}\: \\ $$$${get}\:{y}=\mathrm{0}\:{and}\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:{lim}=\frac{\mathrm{0}}{{x}^{\mathrm{2}} +\mathrm{0}}=\mathrm{0} \\ $$$${get}\:{y}\rightarrow\mathrm{0}\:{and}\:{x}=\mathrm{0}\:\Rightarrow\:{lim}=\frac{{y}^{\mathrm{2}} }{\mathrm{0}+{y}^{\mathrm{2}} }=\mathrm{1} \\ $$$${get}\:\left({y}={x}\right)\rightarrow\mathrm{0}\:\Rightarrow\:{lim}=\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$… \\ $$
Answered by a.lgnaoui last updated on 19/Feb/23
$${I}\:{think}\:{that}\:{a}\:{rule}\:{of}\:{l}\:{hopital} \\ $$$${can}\:{get}\:{a}\:{solutiin} \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:\:\:\:{g}\left({x}\right)={y}^{\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow{a}} \left(\frac{{f}}{{g}}\right)\:={lim}_{{x}\rightarrow{a}} \left(\frac{{f}'}{{g}^{'} }\right) \\ $$$${with}\:\frac{{d}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{{d}\left({x},{y}\right)}=? \\ $$