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Stuck in assignment due tomorrow(in fact this morning)

  1. Sep 21, 2006 #1
    I don't know if there's anyone who bother to answer my questions but still let me try..
    If function f:R^2-->R is f(x,y)=0 if (x,y)=(0,0) and xy^2/(x^2+y^2) otherwise, then prove that f is not continuous at (0,0)
    Supposed to be easy but my brain is dead right now for this..
    it's multivariable analysis question by the way(proof has to be rigorous..)
    Any reply would be appreciated :redface:
     
  2. jcsd
  3. Sep 21, 2006 #2

    StatusX

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    It's sufficient to find a sequence (x_n,y_n) tending to (0,0) but with f(x_n,y_n) not tending to f(0,0)=0. Consider what the function looks like when one of x or y is very small compared to the other.
     
  4. Sep 21, 2006 #3
    Consider what the function looks like when one of x or y is very small compared to the other.

    hmm....
     
  5. Sep 21, 2006 #4
    Got it!!!
    let (x,y)=(1/n_2,1/n) then (x,y)-->(0,0) but f(x,y)-->infinity!! Thanks

    umm... I have 6 more to go...
    next one, if I'm allowed to ask more than 1 question, is
    using same function as above,
    Show directional derivative Dv(f)(x,y) exists for all (x,y) in R_2 and for all vector v in R_2. Then Calculate Dv(f)(x,y).
    Now, first part I'm not quite sure, second part, can I use the fact that
    Dv(f)(x,y)=sum over partial derivatives(with coefficients)
     
  6. Sep 21, 2006 #5

    StatusX

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    Wait, sorry, I think the original function is continuous. Are you sure you copied the question correctly?
     
  7. Sep 21, 2006 #6
    no it's not continuous and I proved it using the method you outlined..?!?!
     
  8. Sep 21, 2006 #7

    StatusX

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    If that's supposed to be [itex](x_n,y_n)=(1/n^2,1/n)[/itex], then [itex]f(x_n,y_n)=(1/n^4)/(1/n^2+1/n^4)=1/(n^2+1) \rightarrow 0[/itex]. Maybe x,y can be complex, or you need to prove it is continuous?
     
    Last edited: Sep 21, 2006
  9. Sep 21, 2006 #8
    umm... its (1/n_4)/(1/n_4!!+1/n_4)=n_2 giving infinity...
    Anyway could you please....... answer my next question plz... I'm sleepy and I need to finish this off plz plz plz
     
  10. Sep 21, 2006 #9

    StatusX

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    Hold on man. I can't help you if I don't understand you. What does n_2 mean? If it's [itex]n^2[/itex], check your algebra again (or make sure you copied the function right in the first post).
     
  11. Sep 21, 2006 #10

    matt grime

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    the symbol _ means subscript. ^ means superscript. and you have a mistake. StatusX is correct when he subs in (n^-2,n^-1), and you are not.
     
  12. Sep 21, 2006 #11
    ohhhhhhhhhhhhhhhhhhhhhhhhhhhhh
    sorry my bad my bad apology*1000000000000000
    I copied down function wrong... it's supposed to be xy^2/(x^2+y^4!!!!)
    my mistake sorry !!
     
  13. Sep 21, 2006 #12
    it's getting very late here and my brain is not functioning well.. sorry about that
     
  14. Sep 21, 2006 #13
    suppose [tex]x = ky^2, y \rightarrow 0[/tex]
    then we know this: [tex]x \rightarrow 0[/tex] also.
    so:
    [tex]\lim \frac{xy^2}{x^2+y^4}= \lim \frac{ky^4}{k^2y^4+y^4}=\lim \frac{k}{k^2+1}[/tex]
    the limit doesn`t exit, so it`s not continous at (0,0).
     
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