# Prove that $g(x,y) = \frac{x^3y}{x^2+y^2}$ with $g(0,0)=0$ is continuous

Consider the function $g:\mathbb{R^2} \rightarrow \mathbb{R}$ defined by

$g(x,y) = \left\{\begin{matrix} \dfrac{x^3y}{x^2+y^2}& \textrm{if } (x,y) \neq (0,0) \\ 0 & \textrm{if }(x,y) = (0,0) \end{matrix}\right.$

How do I prove this is continuous?

I know that I have to:

Consider $x_0 \in \mathbb{R^2}$, and let $\epsilon > 0$ be given. I am confused on how to pick a $\delta$ that will work such that $d(x,x_o) < \delta$.

• It's probably easier to translate it into polar coordinates. You'll see the absolute value of the function is bounded by $r^2$ which goes to zero as $r$ goes to zero. – Gregory Grant Feb 17 '16 at 2:19
• Related questions here and here. – Pedro Feb 17 '16 at 2:32
• My You want the $\epsilon$-$\delta$ approach!! Have you changed your mind? – Mhenni Benghorbal Feb 17 '16 at 3:00

Since $f(x,y)$ for $(x,y)\neq (0,0)$ is a composition of continous functions, it is continuous on $\mathbb{R}^2 \setminus \{(0,0)\}$. For continuity on $\mathbb{R}^2$ we only have to show that the limit

$$\lim_{(x,y)\to (0,0)} f(x,y) = 0$$

exists, e.g. by changing the limit using polar coordinates:

$$x = r \cdot \cos(\theta)$$

$$y = r \cdot \sin(\theta)$$

$$\lim_{(x,y)\to (0,0)} {x^3y\over x^2+y^2} = \lim_{r \to 0} {r^4 \cos^3 \theta \sin \theta \over r^2(\cos^2 \theta + \sin^2 \theta)}= \lim_{r \to 0} {r^2 \cos^3 \theta \sin \theta } = 0$$

Thus, $f(x,y)$ is continuous on $\mathbb{R}^2$.

This function is clearly continuous on $\mathbb{R}^2\setminus\{(0,0)\}$. The only effort will be to prove that it is continuous over $(0,0)$.

Since $\lvert\frac{x^3y}{x^2 + y^2}\rvert = \frac{x^2 \cdot |xy|}{x^2 + y^2} < \frac{(x^2 + y^2) (\frac{x^2 + y^2}{2})}{x^2 + y^2} = \frac{x^2 + y^2}{2}$, the norm of $f(x,y)$ is smaller than the norm of $(x,y)$ thus the continuity of $f$ over $(0,0)$ is guaranteed.

• Nice way to show this using an inequality rather than brute force. – RRL Feb 17 '16 at 3:23

Use the fact that, for $(x,y)\ne(0,0)$, $$\left|\frac{2xy}{x^2+y^2}\right|\le 1$$ Then $$\left|\frac{x^3y}{x^2+y^2}\right|\le\frac{x^2}{2}$$ and the squeeze theorem tells you that $$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^2+y^2}=0$$

$$|g(x, y) | \leq x^2 + y^2 < \epsilon \implies \sqrt{ x^2 + y^2} \leq \sqrt{\epsilon} = \delta$$

• so you are saying that I shold let $\delta = \sqrt{\epsilon }$ ? – johnbowen Feb 17 '16 at 3:44