Proving that $f(x,y) = \frac{x^2y}{x^2+y^2}$ with $ f(0,0)=0$ is continuous

Question

$$f(x,y)=\begin{cases} \dfrac{x^2y}{x^2+y^2}\Leftrightarrow x^2+y^2\not=0\\0\Leftrightarrow x=y=0\end{cases}$$

Answer 1

If $x\neq 0$ we have $$\left|\frac{x^2y}{x^2+y^2}\right|=\frac{|y|}{1+\displaystyle\frac{y^2}{x^2}}\leq |y|$$ So, for all $(x,y)\in\mathbb{R}^2$ we have $0\leq |f(x,y)|\leq|y|$. It follows from the squeeze theorem that $$\lim_{(x,y)\to(0,0)}f(x,y)=0=f(0,0)$$ and thus $f$ is continuous at origin.

If $a\neq 0$ or $b\neq 0$, it follows from basic properties of limits that $$\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b)$$ and thus $f$ is also continuous at any point $(a,b)\in \mathbb{R}^2\setminus\{0\}$.

Answer 2

It is evidently continuous out of the origin. We will inspect the origin.

Let $\epsilon>0$. Take $\delta:=\epsilon$

If $\vert \textbf{x} \vert < \delta$, then $|y| < \delta$ (where $\textbf{x}=(x,y)$). Therefore:

$\displaystyle |f(x,y)| =\frac{|x^2y|}{|x^2+y^2|}< |y|<\epsilon$

Answer 3

The function is continuous at $(0,0)$ because $\|\frac{x^2y}{x^2+y^2}\|_2=\|\frac{x^2y+y^3-y^3\|}{\|(x,y)\|_2^2}\leq\|(0,y)\|+ \|\frac{(0,y^3)\|}{\|(x,y)\|^2}\leq \|(x,y)\|+\frac{\|(x,y)\|_2^3}{\|(x,y)\|_2^2}=2\|(x,y)\|_2$ and the rest is corollary.