Question Number 131444 by rs4089 last updated on 04/Feb/21
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Answered by Olaf last updated on 05/Feb/21
![polar coordinates : ((xy+2)/(x^2 +y^2 )) = ((rcosθ.rsinθ+2)/(r^2 cos^2 θ+r^2 sin^2 θ)) = (((1/2)r^2 sin(2θ)+2)/r^2 ) = ((sin(2θ))/2)+(2/r^2 ) lim_(r→0) ((xy+2)/(x^2 +y^2 )) = +∞](https://www.tinkutara.com/question/Q131460.png)
$$\mathrm{polar}\:\mathrm{coordinates}\:: \\ $$$$\frac{{xy}+\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\:\frac{{r}\mathrm{cos}\theta.{r}\mathrm{sin}\theta+\mathrm{2}}{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta+{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{2}\theta\right)+\mathrm{2}}{{r}^{\mathrm{2}} }\:=\:\frac{\mathrm{sin}\left(\mathrm{2}\theta\right)}{\mathrm{2}}+\frac{\mathrm{2}}{{r}^{\mathrm{2}} } \\ $$$$\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{xy}+\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\:+\infty \\ $$