# Thread: f is not continuous

1. ## f is not continuous

Hello everyone

$f: \mathbb{R}^2 \to \mathbb{R}, \ \ \ x,y \in \mathbb{R}$

$f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, & \mbox{if } (x,y) \not= (0,0) \\ 0, & \mbox{if } (x,y) = (0,0) \end{cases}$

Show that f is not continuous in (0, 0)

I really do not know how to solve that.

Rapha

2. Hello,
Originally Posted by Rapha
Hello everyone

$f: \mathbb{R}^2 \to \mathbb{R}, \ \ \ x,y \in \mathbb{R}$

$f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, & \mbox{if } (x,y) \not= (0,0) \\ 0, & \mbox{if } (x,y) = (0,0) \end{cases}$

Show that f is not continuous in (0, 0)

I really do not know how to solve that.

Rapha
In order to show it is continuous, you must show that $\lim_{(x,y) \to (0,0)} \frac{xy}{x^2+y^2}=0$, for any direction.
Since you want to show it is not continuous, it is sufficient to find a direction for which the limit is not 0.

Switch to polar coordinates. The limit will become :
$\lim_{r \to 0} \frac{r^2 \cos \theta \sin \theta}{r^2}=\lim_{r \to 0} \cos \theta \sin \theta$

which is not always equal to 0.

Hence it is not continuous at (0,0)

3. Try two different paths to (0,0). Say $y=x$ and $y=-x$.
Do you get different limits?

4. Originally Posted by Moo
Hello,

...
Merci beaucoup.

Originally Posted by Plato
Try two different paths to (0,0). Say $y=x$ and $y=-x$.
Do you get different limits?
Thank you very much. I get different limits. Good stuff :-)