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Question Number 127009 by BHOOPENDRA last updated on 26/Dec/20

$$f(x,y)=\{\frac{2x(x^2-y^2)}{x^2+y^2},(x,y)\ne (0,0)$$$$\{0,(x,y)=(0,0)$$$$checkcontinuity$$

Answered by Olaf last updated on 26/Dec/20

$$\mathrm{Let}x=r\cos t\mathrm{and}y=r\sin t$$$$\left(\mathrm{polar}\mathrm{coordinates}\right)$$$$\Rightarrow f(r,t)=\frac{2r\cos t.r^2({\cos }^{2}t-{\sin }^{2}t)}{r^2}$$$$f(r,t)=2r\cos t\cos \left(2t\right)$$$$\mathrm{\forall }t\left(\mathrm{for}\mathrm{any}\mathrm{direction}\right),\underset{r\to 0}{\lim }f(r,t)=0=f(0,0)$$$$f\mathrm{is}\mathrm{continuous}\mathrm{in}(x,y)=(0,0).$$

Commented by BHOOPENDRA last updated on 26/Dec/20

$$thankssir$$

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