# Continuity of function in $\mathbb{R}^2$

Where should I start if I want to study the continuity of a function in $$\mathbb{R}^2$$? Like this one: $$f(x,y)= \begin{cases} \frac{x^2y}{x^2+\sqrt{y}} & \quad\text{if } y>0,\\ 0 & \quad\text{otherwise.}\\ \end{cases}$$ I think $$f$$ is continuous except at $$(0,0)$$, so I have to take the limit to see if it's continuous, right? But I'm confused about the piecewise function, which would be $$0$$ at that point... Could someone help me?

You can notice that for $$y \gt 0$$

\begin{aligned}\vert f(x,y) \vert &= \left\vert \frac{x^2y}{x^2+\sqrt{y}} \right\vert\\ &\le \left\vert \frac{x^2y}{x^2} \right\vert = \vert y \vert \end{aligned} and that the exact same inequality is also satisfied for $$y \le 0$$ as in that case $$f$$ vanishes.

As for any $$u \in \mathbb R$$ $$\lim\limits_{(x,y) \to (u,0)} \vert y \vert = 0$$, you can conclude that $$f$$ is continuous at $$(u,0)$$ as $$f(u,0) = 0$$.

Also, $$f$$ is continuous at $$(u,v)$$ with $$v \neq 0$$ using continuity of composition of continuous maps.

Finally, $$f$$ is continuous everywhere.

Continuity at (0,0);

Let $$\epsilon >0$$ be given.

Let $$\delta=\epsilon,$$ and $$(x, y) \not =(0, 0);$$

$$0\le |f(x,y) - 0| \le |\dfrac{x^2|y|}{x^2+\sqrt{|y|}}| \le$$

$$|y|\le \sqrt{x^2+y^2} \lt \delta=\epsilon.$$

For $$y=0$$ we have $$f(x, 0)=0$$ and trivially

$$0=|f(x,0)-0| \lt \delta =\epsilon$$.