What is the domain of $f(x,y) = xy/\sqrt{x^2+y^2}$

Question

I think the domain is all real except $(0,0)$, because this point anulates the denominator, but what about with the fact that the limit exists when $(x,y)$ tends to $(0,0)$?

Answer 1

The domain of $f$ is all real numbers except when $\sqrt{x^2+y^2} = 0$. But, this occurs when $x^2 + y^2 = 0$ which, in turn, occurs only when $x=y=0$. Therefore,

$$ \text{Domain}(f) = \mathbb{R}^2 \setminus \{ (0,0) \} $$

Answer 2

It can be made continuous, with domain $\mathbb{R}^2$, by defining $f(0,0)=0$, but the domain of the given function excludes $(0,0)$.