continuity multiple variable functions

Question

we just started learning about multiple variable functions and I really don't know how to solve this one:

$$f(x,y) = \begin{cases} x^2y & \text{if x$\in$$\Bbb Q$} \\ y & \text{if if x$\notin$$\Bbb Q$} \end{cases}$$

Is $f$ continuous at $(0, 0)$? and $(0,1)$?

This function looks similar to Dirichlet function, so I think that it is continuous only when $x=y=0$, or when $x=1$ and $y$ can be anything. I really don't know how to formally calculate this and would really use some help.

Thank you!

Answer 1

For $(0,1)$.

put $$z_n=(x_n,y_n)=(\frac{\sqrt{2}}{n},1)$$ then we have $$\lim_{n\to+\infty}z_n=(0,1)$$ $$f(z_n)=1$$ $$f(0,1)=0$$

can you conclude.

Answer 2

For $(0,0)$: $f(0,0)=0$ and, if $(x,y)$ is close to $(0,0)$, then $\bigl|f(x,y)\bigr|\leqslant|y|$ and you can use the squeeze theorem.

Answer 3

At $(0,0)$ it is continuous indeed for any $\epsilon>0$, assuming $(x,y)=(\delta_1,\delta_2)$ with $\delta_1,\delta_2<\epsilon$, we have that

$$|f(x,y)-0|\le \max\{\delta_1^2\delta_2,\delta_2\}=\delta_2<\epsilon$$

For $(0,1)$ we need to show that te limit doesn't exist.