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Question Number 122623 by Ar Brandon last updated on 18/Nov/20

$$\mathrm{Find}\mathrm{the}\mathrm{limits}\mathrm{at}(0,0)\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{functions}:$$$$1.f(\mathrm{x},\mathrm{y})=\frac{{\mathrm{x}}^2{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}2.f(\mathrm{x},\mathrm{y})=\frac{\mathrm{xy}}{{\mathrm{x}}^2+{\mathrm{y}}^2}3.f(\mathrm{x},\mathrm{y})=\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}}$$$$4.f(\mathrm{x},\mathrm{y})=\frac{{\mathrm{x}}^2-{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}5.f(\mathrm{x},\mathrm{y})=\frac{{3\mathrm{x}}^2+\mathrm{xy}}{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}6.f(\mathrm{x},\mathrm{y})=\frac{\mathrm{x}+\mathrm{y}}{{\mathrm{x}}^2+{\mathrm{y}}^2}$$$$7.f(\mathrm{x},\mathrm{y})=\frac{1+{\mathrm{x}}^2+{\mathrm{y}}^2}{\mathrm{y}}\sin \left(\mathrm{y}\right)8.f(\mathrm{x},\mathrm{y})=\frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{{\mathrm{x}}^2+{\mathrm{y}}^2}$$

Answered by mathmax by abdo last updated on 18/Nov/20

$$1)\mid \mathrm{xy}\mid ⩽\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{2}\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2⩽\frac{1}{4}{({\mathrm{x}}^2+{\mathrm{y}}^2)}^2\Rightarrow \frac{{\mathrm{x}}^2{\mathrm{y}}^2}{{\mathrm{x}}^2+{\mathrm{y}}^2}⩽\frac{1}{4}({\mathrm{x}}^2+{\mathrm{y}}^2)\mathrm{wehave}$$$${\lim }_{(\mathrm{x},\mathrm{y})\to (0,0)}\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{4}=0\Rightarrow {\lim }_{(\mathrm{x},\mathrm{y})\to (0,0)}\mathrm{f}(\mathrm{x},\mathrm{y})=0$$$$2){\lim }_{\mathrm{x}\to \mathrm{o}}\mathrm{f}(\mathrm{x},\mathrm{x})={\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{x}}^2}{{2\mathrm{x}}^2}=\frac{1}{2}$$$$\mathrm{let}\mathrm{x}=\mathrm{rcos}\theta \mathrm{and}\mathrm{y}=\sin \theta \mathrm{we}\mathrm{have}$$$$\mathrm{f}(\mathrm{x},\mathrm{y})=\frac{{\mathrm{r}}^2\cos \theta \sin \theta }{{\mathrm{r}}^2}=\frac{1}{2}\sin \left(2\theta \right)\Rightarrow {\lim }_{(\mathrm{x},\mathrm{y})}\mathrm{f}(\mathrm{x},\mathrm{y})\mathrm{dont}\mathrm{exist}..!$$

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