Existence of a two variable function limit

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, so that the limit does not exist along y = x.Sorry about that.
  • #1
Jalo
120
0

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.
 
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  • #2
Jalo said:

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
Jalo said:

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
SammyS said:
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 ?
 
Last edited:
  • #5
Jalo said:
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
Jalo said:
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
Jalo said:
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
SammyS said:
So, there you have it. The limit is +∞ along one path,

I would disagree with that. See my post #7.
 
  • #9
LCKurtz said:
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.
 

1. What is the definition of a limit for a two variable function?

The limit of a two variable function is a value that the function approaches as the two variables get closer and closer to a particular point. It represents the behavior of the function at that specific point.

2. How is the limit of a two variable function calculated?

The limit of a two variable function can be calculated by approaching the point of interest along different paths and evaluating the function at each point. If the values approach the same number, then that number is the limit.

3. Can a two variable function have a limit at a point where it is not defined?

Yes, a two variable function can have a limit at a point where it is not defined. This is because the limit only considers the behavior of the function near the point of interest, not necessarily at that exact point.

4. What is the difference between a one-sided and two-sided limit for a two variable function?

A one-sided limit only considers the behavior of the function when approaching the point of interest from one direction, while a two-sided limit considers the behavior from both directions. In other words, a two-sided limit requires the function to approach the same number from both the left and right sides.

5. Can the limit of a two variable function be different depending on the path of approach?

Yes, the limit of a two variable function can be different depending on the path of approach. This is because the behavior of the function can vary along different paths, and the limit only considers the behavior near the point of interest.

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