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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?

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1  
It seems like another instance Wolfram|Alpha is wrong. –  Clayton Feb 19 '13 at 6:41
1  
WA is wrong. ${}{}{}{}$ –  copper.hat Feb 19 '13 at 6:50
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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
10  
Nooooooooooooo! –  in_wolframAlpha_we_trust Feb 19 '13 at 9:05
1  
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

4 Answers 4

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

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

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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.

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1  
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

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

follows by the squeezing theorem.

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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

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