Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
It seems like another instance Wolfram|Alpha is wrong. – Clayton Feb 19 '13 at 6:41
WA is wrong. ${}{}{}{}$ – copper.hat Feb 19 '13 at 6:50
Remember that you can point out the error to Wolfram|Alpha. At the bottom of every query, there's a link to send the company feedback. – Kevin Feb 19 '13 at 7:20
Nooooooooooooo! – in_wolframAlpha_we_trust Feb 19 '13 at 9:05
When I run the query on W|A, I get a message saying "Standard computation time exceeded..." Might that have a bearing on the incorrect answer? – robjohn Feb 19 '13 at 11:56
up vote 8 down vote accepted

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. $$ \lim_{(x,y)\to(0,0)}\frac{xy}{|x|+|y|}=0.$$

share|cite|improve this answer

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, $$ \lim_{(x,y)\to(0,0)}\left|\frac{xy}{|x|+|y|}\right|\le\lim_{(x,y)\to(0,0)}\frac{\max(|x|,|y|)}{2}=0\tag{4} $$

share|cite|improve this answer

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.

share|cite|improve this answer
Like your choice of words. bleh blah – Matsemann Feb 19 '13 at 9:54
Thanks! I edited the OP. – Pablo Montepagano Feb 19 '13 at 13:36

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


follows by the squeezing theorem.

share|cite|improve this answer
This can be extended to show that $$ \left|\frac{xy}{|x|+|y|}\right|\le\min(x,y) $$ a nice complement to my answer :-) – robjohn Feb 19 '13 at 13:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.