What is a good approach to proving $f$ is continuous in $\bar 0$

Question

I'm reading up on metric spaces, and the following question appeared in the continuous functions section:

$$\text{Let} f:\mathbb R^2 \to \mathbb R \text{ be defined such that } f(\bar0)=0 \text{ and} \\ f(x,y) = \frac{xy^2}{x^2+y^2} \text{, for } (x,y) \not = \bar 0. \ \text{Show that } f \text{ is continuous in the origin.}$$

I haven't studied vector calculus, so the continuity of two-dimensional functions is unclear to me. If I've understood correctly, the definition in this case is $|f(x,y)| < \epsilon$ for $|\sqrt{x^2 + y^2}| < \delta$. However, I can't see the desired simplifications and a choice for $\delta$. Also, since I'm studying topology, is there a "more topologically" relevant way to answer this question (i.e an argument based on open sets)?

Answer 1

We have $$ \left|\frac{xy^2}{x^2 + y^2}\right|=|x|\cdot\left|\frac{y^2}{x^2 + y^2}\right|\leq |x|\cdot \left|\frac{y^2}{y^2}\right| = |x| $$ Ok, this is nicely simplified. We can now start our actual proof.

Let $\varepsilon>0$, and pick^* $\delta = \varepsilon$. Assuming $\sqrt{x^2 + y^2}<\delta$, we get $$ \left|\frac{xy^2}{x^2 + y^2}\right|\leq |x|=\sqrt{x^2}\leq \sqrt{x^2 + y^2}<\delta = \varepsilon $$ showing that our function is continuous at the origin

^* I fill in $\delta = \varepsilon$ after actully doing the calculations further down, so that I know what I need. But I write it before the calculations in the finished proof, so it looks like I pulled it out of thin air, somehow knowing that it would work out.

Answer 2

With polar coordinates $x= r \cos t, y=r \sin t$ we see that

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

Hence, if $ \epsilon >0$ is given, let $ \delta= \epsilon.$