1. Not finding help here? Sign up for a free 30min 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!

Multivariable Calculus Limit process

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine whether the limit exists; if it does, what is it?

    2. Relevant equations

    take the limit as (x,y) -> (0,0) of f(x,y) where f(x,y) = (x^6-y^6)/(x^3-y^3)

    3. The attempt at a solution

    What i started doing was approaching along the line y=0 and that would give
    lim as (x,0) -> (0,0) x^6/x^3 simplifying to the lim as (x,0) -> (0,0) x^3 which is 0.

    Then I approached along the line x=0 and that would give
    lim as (0,y) -> (0,0) -y^6/-y^3 simplifying to the lim as (0,y) -> (0,0) y^3 which is 0.

    Then i approached along the line y=x
    lim as (x,x) -> (0,0) (x^6-x^6)/(x^3-x^3) = 0/0 which is an indeterminate form so I use L'Hospital's rule a few times and continue to get 0/0. This rational function is the same as saying (x-x)/(x-x).

    I conclude that the limit does not exist becuase it can not be evaluated any further. Am i correct? Is there something else I shoudl be doing? This homework problem is due March 22.
    Thank you for your help.
     
  2. jcsd
  3. Mar 5, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi theno1katzman! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    hmm … I've a nasty feeling that the domain of the function isn't supposed to include the line x = y, in which case you can't approach along that line.

    Try approaching along a curve on which f(x,y) is constant. :wink:
     
  4. Mar 5, 2010 #3
    Thanks Tim, when I tried approaching (0,0) along the curve y=e^x I will get
    limit of (x,e^x)->(0,0) of (x^6-e^(6 x))/(x^3-e^(3 x)) = 1

    This is a different answer for the limit above which was 0, therefore the limit of the original function as (x,y) approaches (0,0) does not exist.
     
  5. Mar 6, 2010 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (please use the X2 tag just above the Reply box :wink:)

    erm :redface: … noooo …

    y = ex doesn't get anywhere near (0,0), does it? :wink:

    Try again, to find a curve on which f(x,y) is constant (you may have to solve a quadratic). :smile:
     
    Last edited: Mar 6, 2010
  6. Mar 6, 2010 #5
    Let me know if this is correct. If I approached (0,0) along the path y=x n where n is any positive integer for example y=x2 then after factoring I would get

    lim (x,x2)-> (0,0) x3(x2-x+1)/(x2+x+1) = 0/1 = 0.

    Therefore the limit exists and is 0.
     
  7. Mar 6, 2010 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    I didn't follow that … for y = x2, it would be x3(1 - x6)/(1 - x3), which does --> 0.

    But that doesn't prove it for all paths, does it?

    Try my previous hint, to find a curve on which f(x,y) is constant.
     
  8. Mar 6, 2010 #7
    [tex]y = x^6 + c [/tex] [tex] lim (x,y) -> (0,0) f(x,y) [/tex]
     
  9. Mar 6, 2010 #8
    Did you notice that the numerator is a difference of squares ?
     
  10. Mar 6, 2010 #9
    Ahh yes I will simplify this to [tex]x^3+y^3[/tex] and work the problem again
     
  11. Mar 6, 2010 #10
    So the limit = ?

    The polar coordinates works also :)
     
  12. Mar 6, 2010 #11
    limit would be 0 if I use polar coordinates, question, would polar coordinates prove the limit exists for all paths?
     
  13. Mar 6, 2010 #12
    Yes.
    There is a big difference in solving multivariable limits between the 2-path rule (which is y=mx .. etc) and the polar coordinates method.
    2-path rule proves only that the limit D.N.E and does not prove the existence of the limit.
    Polar coordinates method proves both cases.
     
  14. Mar 6, 2010 #13
    Thank you, I wish my professor would have clarified that in class.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Multivariable Calculus Limit process
Loading...