limit of $\frac{xy}{|x|+|y|}$ at $(0, 0)$ Let $f(x, y)=\frac{xy}{|x|+|y|}$ for $(x, y)\neq (0, 0)$, find $\lim_{(x, y)\to (0, 0)}\frac{xy}{|x|+|y|}$.
What I've tried
let $x= cos\theta, y=sin\theta$, then $f(x,y)=\frac{cos\theta sin\theta}{|cos\theta|+|sin\theta|}$, but this doesn't seems to go anywhere.
And also I tried let $|(x, y)|=P$, then $|xy|\leq P^2$, still not work. Any help, thanks!
 A: Note, for $x \neq 0$, you have that
$$0 \le \left|\frac{xy}{|x| + |y|}\right| \le \left|\frac{xy}{|x|}\right| = |y| \tag{1}\label{eq1A}$$
Thus, since as $(x,y) \to (0,0)$ you have $|y| \to 0$, the squeeze theorem shows that
$$\lim_{(x, y)\to (0, 0)}\frac{xy}{|x|+|y|} = 0 \tag{2}\label{eq2A}$$
As angryavian's comment indicates, with $x = 0$, you can switch the roles of $x$ and $y$ above and then proceed with a similar argument. Alternatively, with $x = 0$ with $y \neq 0$, or $y = 0$ with $x \neq 0$, you can just use that $\frac{xy}{|x| + |y|} = 0$.
Update: One other method you can use, for $x \neq 0$ and $y \neq 0$, is the inequality of arithmetic and geometric means giving that
$$|x| + |y| \ge 2\sqrt{|xy|} \tag{3}\label{eq3A}$$
This then gives
$$0 \le \left|\frac{xy}{|x| + |y|}\right| \le \frac{|xy|}{2\sqrt{|xy|}} = \frac{\sqrt{|xy|}}{2} \tag{4}\label{eq4A}$$
where, similar to before, as $(x,y) \to (0,0)$, you have $\frac{\sqrt{|xy|}}{2} \to 0$.
A: Fleshing out my comment on John Omielan's answer:
$$\left|\frac{xy}{|x|+|y|}\right| \le \frac{|xy|}{\max\{|x|, |y|\}} \le \min\{|x|, |y|\}$$
which tends to zero as $(x,y) \to (0,0)$.
A: Your attempt using polar coordinates was almost there.   You just inadvertently dropped a factor of $r$ in the numerator.
Note that for $r\ne0$
$$\begin{align}
 \frac{xy}{|x|+|y|}&=\frac{r^2 \cos(\theta)\sin(\theta)}{r\left(|\cos(\theta)|+|\sin(\theta)| \right)}\\\\
&=\frac{r\sin(\theta)\cos(\theta)}{|\sin(\theta)|+|\cos(\theta)|}
\end{align}$$
Clearly this goes to $0$ as $r\to0$.
