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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?

Martin Sleziak
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Pablo M
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    It seems like another instance Wolfram|Alpha is wrong. – Clayton Feb 19 '13 at 06:41
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    WA is wrong. ${}{}{}{}$ – copper.hat Feb 19 '13 at 06:50
  • Note that it says "value may depend on $x,y$ path in complex space" -- it's not quite clear whether it's considering $(x,y)$ as a complex number or both $x$ and $y$ as complex numbers, but it doesn't matter since most of the proofs in the answers also work for complex $x$ and $y$. – joriki Feb 19 '13 at 06:51
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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 07:20
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    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
  • A couple of years later and... they've fixed it! http://www.wolframalpha.com/input/?i=lim+when+(x,y)-%3E(0,0)+of+xy%2F(%7Cx%7C%2B%7Cy%7C) – Pablo M Apr 23 '18 at 00:27

6 Answers6

19

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

robjohn
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12

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

Hagen von Eitzen
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9

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.

Emanuele Paolini
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7

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.

egreg
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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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Hint: By the arithmetic-geometric inequality $$ \frac{|xy|}{|x|+|y|}\leq\frac{\sqrt{|xy|}}{2}. $$

3

Any function $f(x,y)$ such that $|f(x,y)|\le |x|$ has limit $0$ at $(0,0).$ And we have that here:

$$\left | \frac{xy}{|x|+|y|} \right | = |x|\left | \frac{|y|}{|x|+|y|} \right | \le |x|.$$

zhw.
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