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Question Number 191733 by Spillover last updated on 29/Apr/23
$${Show}\:{that}\: \\ $$$$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\:\:\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:{does}\:{not}\:{exist} \\ $$
Answered by mehdee42 last updated on 30/Apr/23
$${let}\:{y}={x}\rightarrow\:{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} \:\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{0} \\ $$$${let}\:{y}=\mathrm{2}{x}\rightarrow\:{lim}_{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} \:\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${therefor}\:\:{this}\:{limit}\:{does}\:{not}\:{exist} \\ $$$$ \\ $$$$\: \\ $$
Answered by Frix last updated on 29/Apr/23
$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{{r}\mathrm{cos}\:\theta}\\{{r}\mathrm{sin}\:\theta}\end{pmatrix} \\ $$$$\underset{\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\rightarrow\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\:−{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\:+{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta}\:= \\ $$$$=\mathrm{cos}\:\mathrm{2}\theta\:\Rightarrow\:\mathrm{the}\:\mathrm{limit}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}\:\mathrm{as}\:\mathrm{its} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{value}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{direction} \\ $$
Commented by Spillover last updated on 30/Apr/23
$${thank}\:{you} \\ $$