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Question Number 142462 by BHOOPENDRA last updated on 01/Jun/21

Answered by MJS_new last updated on 01/Jun/21

$$y=px\Rightarrow \frac{xy+2}{x^2+y^2}=\frac{p}{p^2+1}+\frac{2}{p^2+1}\times \frac{1}{x^2}$$$$p\in \mathbb{R}\Rightarrow \frac{p}{p^2+1}=q\in \mathbb{R}\wedge r=\frac{2}{p^2+1}\in {\mathbb{R}}^{+};(\wedge r\ne 0)$$$$\Rightarrow$$$$\underset{x\to 0;y\to 0}{\lim }\frac{xy+2}{x^2+y^2}=q+\underset{x\to 0}{\lim }\frac{r}{x^2}=+\mathrm{\infty }$$

Commented by BHOOPENDRA last updated on 01/Jun/21

$$txsir$$

Answered by mathmax by abdo last updated on 01/Jun/21

$$\mathrm{x}=\mathrm{rcos}\theta \mathrm{and}\mathrm{y}=\mathrm{rsin}\theta \Rightarrow {\lim }_{(\mathrm{x},\mathrm{y})\to (0,0)}\frac{\mathrm{xy}+2}{{\mathrm{x}}^2+{\mathrm{y}}^2}$$$$={\lim }_{\mathrm{r}\to 0}\frac{{\mathrm{r}}^2\cos \theta \sin \theta +2}{{\mathrm{r}}^2}=\cos \theta \sin \theta +{\lim }_{\mathrm{r}\to 0}\frac{2}{{\mathrm{r}}^2}=+\mathrm{\infty }$$

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