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Existence of a two variable function limit

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  • #1
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Homework Statement


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)


Homework Equations





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.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


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)


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.
Try along lines like ##y=mx## for various ##m##.
 
  • #3
SammyS
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Homework Statement


In this exercise I'm asked to find out if the limit of the function f(x,y) exists.

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

Homework Equations



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.
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
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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.
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 ?
 
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  • #5
HallsofIvy
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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)
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.

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 ?
 
  • #6
SammyS
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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)
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 +∞.
As to x=y³

sqrt(y³*y) / ( (y³)² - y²) = y² / (y⁶ - y²) = 1 / (y⁴ - 1) = -1
So, there you have it. The limit is +∞ along one path, and -1 along another.
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 ?
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
LCKurtz
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Could someone just comment on the result of the x=y substitution? Have I done it incorrectly?
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
LCKurtz
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So, there you have it. The limit is +∞ along one path,
I would disagree with that. See my post #7.
 
  • #9
SammyS
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I would disagree with that. See my post #7.
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.
 

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