Answer to Question #170318 in Real Analysis for Prathibha Rose

Question

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)

Answer 1

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

$\lim_{x,y\to 0} \frac{xy}{x^2+y^2}=\lim_{r\to 0} \frac{r\cos{\theta}\text{ } r\sin{\theta}}{r^2}=\lim_{r\to 0}\sin\theta \cos\theta$

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