# Continuity of a function $f: \mathbb{R}^2 \to \mathbb{R}$

It's easy to check that the function

$$f_1(x, y) = \begin{cases}\frac{x y}{x^2 + y^2} &\text{if (x, y) ≠ (0, 0)}\\0&\text{if (x, y) = (0, 0)}\end{cases}$$

is not continuous in $0$, because e.g. the sequence $f(\frac{1}{n}, \frac{1}{n})_{n \in \mathbb{N}}$ won't converge against $0$.

But how can I prove that the very similar function $f_2$, defined as:

$$f_2(x, y) = \begin{cases}\frac{x y^2}{x^2 + y^2} &\text{if (x, y) ≠ (0, 0)}\\0&\text{if (x, y) = (0, 0)}\end{cases}$$

actually is continuous in $0$? Choosing any sequence of $\mathbb{R}^2$ that converges against $(0, 0)$ didn't lead anywhere so far, and I imagine that using the $\epsilon-\delta$-criteria could maybe work, but would probably get to complicated and that there's hopefully an easier way to check this.

• Is this the integral of a function? If so then it is obviously continuous by the FTC – Squirtle Jun 10 '15 at 16:59
• Hint: use polar coordinates. – TZakrevskiy Jun 10 '15 at 17:02

$$\left|\, f(x,y) - f(0,0) \, \right| = \left| \frac{xy^2}{x^2 + y^2} \right| = |\,r \cos\theta\sin^2\theta\, | \leq r$$
Hence given $\epsilon > 0$, choose $\delta = \epsilon$. Then
$$\left|(x,y) - (0,0)\right| = r < \delta \ \Longrightarrow \ \left|\, f(x,y) - f(0,0) \, \right| < \epsilon$$
If you put $x=\rho\cos\left(\theta\right)$ and $y=\rho\sin\left(\theta\right)$ you get $$\lim_{\left(x,y\right)\rightarrow\left(0,0\right)}\frac{xy^{2}}{x^{2}+y^{2}}=\lim_{\rho\rightarrow0^{+}}\frac{\rho^{3}\cos\left(\theta\right)\sin^{2}\left(\theta\right)}{\rho^{2}}=0.$$
Recall $|xy|\le (x^2+y^2)/2.$ Thus $|f(x,y)| \le |y|/2,$ which gives the result.