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

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


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