Answer to Question #170318 in Real Analysis for Prathibha Rose

Question #170318

Prove that the function f:R2 to R defined by f(x,y) =

{ xy/(x2+y2) if (x,y) not equal to (0,0)

0, if (x,y) = (0,0)

Is not continuous at (0,0)


1
Expert's answer
2021-03-17T09:47:10-0400

To verify whether this function is continuous at (0;0)(0;0) we need to study the limit of xyx2+y2\frac{xy}{x^2+y^2} when (x,y)(0,0)(x,y)\to(0,0). When calculating limits of this type (especially in R2\mathbb{R}^2) , it is very useful to pass into the polar coordinate system, where the limit (x,y)(0,0)(x,y)\to (0,0) is replaced by a limit r0r\to 0 independently of θ\theta. In polar coordinates x=rcosθ,y=rsinθx=r\cos \theta, y=r\sin \theta :

limx,y0xyx2+y2=limr0rcosθ rsinθr2=limr0sinθcosθ\lim_{x,y\to 0} \frac{xy}{x^2+y^2}=\lim_{r\to 0} \frac{r\cos{\theta}\text{ } r\sin{\theta}}{r^2}=\lim_{r\to 0}\sin\theta \cos\theta

And we see that this limit depends on θ\theta and thus the limit r0r\to 0 does not exist, so the function f(x,y)f(x,y) is not continuous at (0,0)(0,0). For more visibility, we can calculate the limit along x=0,y0x=0, y\neq 0 which gives f(0,y)=00f(0,y)=0\to 0 and the limit along x=y0x=y\neq 0 which gives f(t,t)=t2/(2t2)=1/21/20f(t,t)=t^2/(2t^2)=1/2\to 1/2\neq 0.


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