How to prove that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is continuous on $\mathbb{R^2} \smallsetminus \{(0, 0)\}$? In a small part of a school assignment I have to prove that $f$ is continuous on $\mathbb{R^2} \smallsetminus \{(0, 0)\}$:
$$f(x, y)=\frac{2xy}{x^2 + y^2}$$
I think by using the quotient rule I can prove this, since the quotient of 2 continuous functions is also a continuous function. As long as the denominator is not zero!
As silly it may be sounds, I get stuck proving that these separate functions $2xy$ and $x^2 + y^2$ are continuous functions. It is obvious that they are continuous, but I don't know how to write a good prove of functions with 2 different variables.
 A: Taking any point not equal to $(0,0)$ you have $x^2+y^2$ continuous and not zero in it. Then knowing $2xy$ is continuous everywhere you divide last on former and get continuous ratio on $\mathbb{R}^2 \setminus (0,0)$.
Answering comment about $2xy$ we can consider it as product of continuous functions $f(x,y)=x, g(x,y)=y, h(x,y)=2$. Same trick works for $x^2+y^2$.
Now about continuity of $f(x,y)=x$ as function $\mathbb{R}^2 \to \mathbb{R}$: inequality $|x-x_0|\leqslant \sqrt{(x-x_0)^2+(y-y_0)^2}$ gives continuity in   any $(x_0,y_0)$, as in definition of continuity we can simply consider $\varepsilon=\delta$.
A: First: proving that $f(x, y) = x^2 + y^2$ is continuous. Let $(a, b)$ and $\epsilon > 0$ be arbitrary. Then $|x^2 + y^2 - a^2 - b^2| = |(x + a)(x - a) + (y - b)(y + b)| \leq |x - a||x - b| + |y - b||y + b|$ by the triangle inequality. Take $(x, y)$ closer than $\delta$ to $(a, b)$. Then $|x - a|, |y - b| < \delta$, which means that also $|x + a| < \delta + 2|a|$, $|y + b| < \delta + 2|b|$ (by the triangle inequality again). So
$$|f(x, y) - f(a, b)| \leq \delta(2\delta + 2|a| + 2|b|)$$
Can you pick a $\delta$ depending only on $a, b, \epsilon$ to make this smaller than $\epsilon$?
For $2xy$, just use the product rule and that projections are continuous.
Actually, now that I think of it the product and sum rule are enough to prove that $x^2 + y^2$ is continuous as well.
