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Question Number 192138 by Mastermind last updated on 09/May/23

$$\mathrm{show}\mathrm{that}$$$$\mathrm{f}(\mathrm{x},\mathrm{y})=\{{}_{0(\mathrm{x},\mathrm{y})=(0,0)}^{\frac{{\mathrm{x}}^2\mathrm{y}}{{\mathrm{x}}^6+{2\mathrm{y}}^2}(\mathrm{x},\mathrm{y})\ne (0,0)}$$$$\mathrm{has}\mathrm{a}\mathrm{directional}\mathrm{derivative}\mathrm{in}\mathrm{the}$$$$\mathrm{direction}\mathrm{of}\mathrm{an}\mathrm{arbitrary}\mathrm{unit}\mathrm{vector}$$$$\varphi \mathrm{at}(0,0),\mathrm{but}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{continous}\mathrm{at}(0,0)$$

Answered by gatocomcirrose last updated on 09/May/23

$$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(0,0)=\underset{x\to 0}{\lim }\frac{\mathrm{f}(\mathrm{x},0)-\mathrm{f}(0,0)}{\mathrm{x}-0}=0$$$$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}(0,0)=\underset{\mathrm{y}\to 0}{\lim }\frac{\mathrm{f}(0,\mathrm{y})-\mathrm{f}(0,0)}{\mathrm{y}-0}=0$$$$\Rightarrow ▽\mathrm{f}(0,0)=(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(0,0),\frac{\partial \mathrm{f}}{\partial \mathrm{y}}(0,0))=(0,0)$$$$\Rightarrow \mathrm{directional}\mathrm{derivative}=$$$$▽\mathrm{f}(0,0).\varphi =(0,0).\varphi =0$$$$$$$$\underset{(x,\mathrm{y})\to (0,0)}{\lim }\frac{{\mathrm{x}}^2\mathrm{y}}{{\mathrm{x}}^6+{2\mathrm{y}}^2}=\underset{x\to 0}{\lim f}(\mathrm{x},0)=0=\underset{\mathrm{y}\to 0}{\lim f}(0,\mathrm{x})$$$$\stackrel{\mathrm{y}={\mathrm{x}}^2}{=}\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4}{{\mathrm{x}}^6+{2\mathrm{x}}^4}=\underset{x\to 0}{\lim }\frac{1}{{\mathrm{x}}^2+2}=\frac{1}{2},\mathrm{contradiction}$$$$\Rightarrow \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{continuous}\mathrm{at}(0,0)$$

Commented by Mastermind last updated on 14/May/23

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{BOSS}$$

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