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Homework Help: 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

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    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

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    (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

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    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.
     
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