Question Number 131444 by rs4089 last updated on 04/Feb/21
Answered by Olaf last updated on 05/Feb/21
$$\mathrm{polar}\:\mathrm{coordinates}\:: \\ $$$$\frac{{xy}+\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\:\frac{{r}\mathrm{cos}\theta.{r}\mathrm{sin}\theta+\mathrm{2}}{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta+{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2}\theta\right)+\mathrm{2}}{{r}^{\mathrm{2}} }\:=\:\frac{\mathrm{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}+\frac{\mathrm{2}}{{r}^{\mathrm{2}} } \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{xy}+\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\:+\infty \\ $$