# Continuity of a $f(x,y) = -x^2$ when $y\geq 0$, $x^2$ when $y < 0$. [closed]

I am troubling with the following question:

Let $f: \mathbb{R^2} \rightarrow \mathbb{R}$ be given by

$f(x,y)= \left\{ \begin{array}{} -x^2 &y \geq 0 \\ x^2 &y<0 \end{array} \right.$

I have to give the points where the function is continuous and the points where the function is discontinuous. I also have to explain why of course. Help is greatly appreciated!

• What happens at $y=0$. Commented May 26, 2017 at 14:49

Your function is continuous when $y\ne 0$ since $(x,y)\mapsto -x^2$ and $(x,y)\mapsto x^2$ are both continuous everywhere.

Now you can look at

$$\lim_{y\to 0^+} f(x,y)$$

and

$$\lim_{y\to 0^-} f(x,y).$$

What can you conclude from here?

• Thanks! Since $\lim_{y\to 0^+} f(x,y) = -x^2$ and $\lim_{y\to 0^-} f(x,y) = x^2$, I can conclude that $f$ is discontinous when $y=0$. One question, what about $(0,0)$?
– user444389
Commented May 26, 2017 at 15:16
• @MartindeQuincey Absolutely, it is discontinuous at $(x,0)$ if $x\ne 0$, but it is continuous at $(0,0)$. Commented May 26, 2017 at 15:21
• Ah okay. To finalize, $f$ is continuous at $(x,y)$ if $y \neq 0$ and at $(0,0)$. And $f$ is discontinous at $(x,0)$ if $x \neq 0$.
– user444389
Commented May 26, 2017 at 15:30