1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Existence of a two variable function limit

  1. Sep 29, 2012 #1
    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. jcsd
  3. Sep 29, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Try along lines like ##y=mx## for various ##m##.
     
  4. Sep 29, 2012 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  5. Sep 30, 2012 #4
    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
  6. Sep 30, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.

     
  7. Sep 30, 2012 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  8. Sep 30, 2012 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  9. Sep 30, 2012 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would disagree with that. See my post #7.
     
  10. Sep 30, 2012 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook