Wrong Wolfram|Alpha limit? $ f(x,y) = \frac {xy}{|x|+|y|} $ for $(x,y)\to(0,0)$ I have this function:
$$ f(x,y) = \frac {xy}{|x|+|y|} $$
And I want to evaluate it's limit when $$ (x,y) \to (0,0)$$ My guess is that it tends to zero. So, by definition, if:
$$
\forall \varepsilon \gt 0, \exists \delta \gt 0 \diagup \\ 0\lt||(x,y)||\lt \delta , \left|\frac{xy}{|x|+|y|}\right| \lt \varepsilon
$$
Then
$$
\lim_{(x,y)\to(0,0)}\frac {xy}{|x|+|y|} = 0
$$
So:
$$
\left|\frac{xy}{|x|+|y|}\right| = \frac{|xy|}{|x|+|y|}
= \frac{|x||y|}{|x|+|y|} \le 1 |y| \lt \delta
$$
So for any $$\delta \lt \varepsilon$$ the inequality is true. Hence, the limit exists and is equal to zero.
Wolfram|Alpha says that the limit does not exist. Am I wrong or is Wolfram|Alpha wrong?
 A: You are right, wolfram is wrong. It might happen...
Only you should correct your exposition of the definition. You say:
By definition, if blah blah, then bleh bleh

you should say:
By definition, blah blah, if bleh bleh

In fact you prove bleh bleh to have blah blah.
A: The first thing to do when computing this kind of limits is trying to isolate a bounded expression.
Assuming $(x,y)\ne(0,0)$ in what follows, we clearly have
$$
\left|\frac{y}{|x|+|y|}\right|\le 1.
$$
Therefore we can write
$$
-|x|\le\frac{xy}{|x|+|y|}\le |x|
$$
and so
$$\lim_{(x,y)\to(0,0)}\frac{xy}{|x|+|y|}=0$$
follows by the squeezing theorem.
A: Hint: By the arithmetic-geometric inequality
$$
\frac{|xy|}{|x|+|y|}\leq\frac{\sqrt{|xy|}}{2}.
$$
A: Any function $f(x,y)$ such that $|f(x,y)|\le |x|$ has limit $0$ at $(0,0).$ And we have that here:
$$\left | \frac{xy}{|x|+|y|} \right | = |x|\left | \frac{|y|}{|x|+|y|} \right | \le |x|.$$
A: Pretty simply, we have
$$
|xy|=\max(|x|,|y|)\min(|x|,|y|)\tag{1}
$$
and
$$
|x|+|y|\ge2\min(|x|,|y|)\tag{2}
$$
Therefore,
$$
\left|\frac{xy}{|x|+|y|}\right|\le\frac{\max(|x|,|y|)}{2}\tag{3}
$$
Thus,
$$
\lim_{(x,y)\to(0,0)}\left|\frac{xy}{|x|+|y|}\right|\le\lim_{(x,y)\to(0,0)}\frac{\max(|x|,|y|)}{2}=0\tag{4}
$$
A: You are right, though you mix up the direction of proof (by what you write, you literally just show "if the limit exists, then it is $0$").
Given $\epsilon>0$, let $\delta=\epsilon$.
Assume $(x,y)\ne(0,0)$ is a point with $|(x,y)|<\delta$. Then especially $0<r<\delta$ with $r:=\max\{|x|,|y|\}$ and hence 
$$ \left|\frac{xy}{|x|+|y|}\right|=\frac{|x|\cdot|y|}{|x|+|y|}\le \frac{r^2}{r+0}=r<\delta<\epsilon,$$
as was to be shown, i.e. 
$$ \lim_{(x,y)\to(0,0)}\frac{xy}{|x|+|y|}=0.$$
