Homework Help: Existence of a two variable function limit

1. Sep 29, 2012

Jalo

1. The problem statement, all variables and given/known data
In this exercice I'm asked to find out if the limif of the function f(x,y) exists.

lim (x,y)→(0,0) sqrt(x*y) / (x^2 - y^2)

2. Relevant equations

3. The attempt at a solution

I've tried to approach it from different coordinates (y=x, y=0, x=0, y=sqrt(x),...) but I didn't found any limit...

I also tried to prove it didn't exist with the iteration property without luck...

Any hint would be highly appreciated.

Thanks.

2. Sep 29, 2012

LCKurtz

Try along lines like $y=mx$ for various $m$.

3. Sep 29, 2012

SammyS

Staff Emeritus
Along x=0 and y=0, the limit definitely exists.

Along the line y=x, what is true of the denominator, x-y2 ?

See what happens along the curve, x=y3.

4. Sep 30, 2012

Jalo

If I substitute y for x I'll get

sqrt(y*y) / (y² - y²) = y / (y² - y²) = 1 / (y-y)

Which is 1/0, an indetermination. I don't get a physical value therefore I can't conclude nothing from this limit.. ( I think? correct me if I'm wrong)

As to x=y³

sqrt(y³*y) / ( (y³)² - y²) = y² / (y⁶ - y²) = 1 / (y⁴ - 1) = -1

I decided to compare it to
x=4*y³

sqrt(4*y³*y) / ( (4*y³)² - y²) = 2*y² / (16*y⁶ - y²) = 2 / (16*y⁴ - 1) = -2

I can now conclude that the limit doesn't exist.

Could someone just comment on the result of the x=y substitution? Have I done it incorrectly?

Thank you very much.
D.

EDIT: Also, can I substitute x for y+n, where n is some random number belonging |R² ?

Last edited: Sep 30, 2012
5. Sep 30, 2012

HallsofIvy

No, "1/0" is NOT an "indetermination" (better English would be "indeterminate"). "0/0" is an indeterminate. Getting "1/0" is itself enough to conclude that the limit does not exist.

6. Sep 30, 2012

SammyS

Staff Emeritus
You're right in that you have division by zero. You have |y|/0 all along the line y=x. But, that's not an indeterminate form. When taking the limit as y→0 you get a result of +∞.
So, there you have it. The limit is +∞ along one path, and -1 along another.
For x = y+n, for n a non-zero number, that does not pass through the origin.

If you literally mean x=y+(a, b), that's not any defined operation as far as I know.

You originally implied that the limit doesn't exist along x=0 and y=0. Those limits do exist, and they are zero. So, your original paths, along y=x, and y=0 or x=0, are enoough to show that the limit does not exist, but using a path such as x = ay3 or y = ax3 does give nice results.

7. Sep 30, 2012

LCKurtz

Others have commented about your calculation of the x=y substitution. But I would comment about its appropriateness. None of the points on y=x are in the domain of the function so it doesn't make sense to even consider that path when studying the limit in question. But, as you have seen, other paths will do.

8. Sep 30, 2012

LCKurtz

I would disagree with that. See my post #7.

9. Sep 30, 2012

SammyS

Staff Emeritus
Yes, of course you are correct !

Unfortunately, I had a typo in post #3, but that should have said that along the line y=x, what does the denominator x2 - y2 become? Of course, the answer is that the denominator is zero, so that the overall expression is undefined along y = x.