Is the function $f$ continuous at $(0,0)$ 
Is the function $f$ continuous at $(0,0)$?
$f(x, y)$ := $\frac{xy}{|x|+|y|}$, if $(x, y)$ $\not= (0, 0)$
and $f(0, 0) := 0$

My attempt:
$f(x, y)$ = $\frac{xy}{x+y}$ if $(x,y) > (0,0)$ and
$f(x, y)$ = $\frac{xy}{-(x+y)}$ if $(x,y) < (0,0)$
So using polar coordinates:
$f(x, y)$ = $\frac{xy}{x+y}$ = $\frac{rcos\theta sin\theta}{cos\theta + sin\theta}$, thus $|f(x,y)| <= r = {(x^2 + y^2)}^{0.5} $ and thus $f$ is continuous.
$f(x, y)$ = $\frac{xy}{-(x+y)}$ = $\frac{-rcos\theta sin\theta}{cos\theta + sin\theta}$, thus $|f(x,y)| <= - r = {-(x^2 + y^2)}^{0.5} $ and thus $f$ is not continuous.
Is my answer correct?
 A: Using polar coordinates, the expression reduces to
$$r\frac{\cos\theta\sin\theta}{|\cos\theta|+|\sin\theta|},$$ where $r$ tends to $0$ and $\theta$ varies arbitrarily.
As the denominator cannot vanish, the angular factor is finite and the limit is $0$.
A: There is no need for polar coordinates. Simply note that $|xy|\leq\max(|x|,|y|)^2$ and $|x|+|y|\geq\max(|x|,|y|)$, so $|f(x,y)|\leq\max(|x|,|y|)$, which tends to $0$ as $(x,y)\to(0,0)$.
A: If $x=r\cos\theta$ and $y=r\sin\theta$, then\begin{align}\left\lvert\frac{xy}{\lvert x\rvert+\lvert y\rvert}\right\rvert&=\frac{r^2\left\lvert\cos(\theta)\sin(\theta)\right\rvert}{r\bigl(\lvert\cos\theta\rvert+\lvert\sin\theta\rvert\bigr)}\\&\leqslant\frac r2,\end{align}since $\left\lvert\cos(\theta)\sin(\theta)\right\rvert\leqslant\frac12$ and $\cos\theta\rvert+\lvert\sin\theta\rvert\geqslant1$. So, $f$ is continuous at $(0,0)$.
A: $(x,y)\not =(0,0)$;
1) $x\not =0$;
$|\dfrac{xy}{|x|+|y|}| \le |y| \lt \sqrt{x^2+y^2};$
2) $y \not =0$;
$|\dfrac{xy}{|x|+|y|}| \le |x| \lt \sqrt{x^2+y^2};$
$\epsilon >0$ given.
Choose $\delta =\epsilon$.
