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Totally differentiable?

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    For any (x,y) other than (0,0).
    [tex]f(x,y)=\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)[/tex]
    For (x,y) = (0,0)
    f(x,y) = 0
    Is f totally differentiable?

    2. Relevant equations

    3. The attempt at a solution
    If the function is not continuous, it can't be differentiable.
    [tex]\lim_{(x,y)\rightarrow(0,0)}\tan(x·y)\sin\left(\frac{1}{x^2+y^2}\right)=\lim_{(x,y)\rightarrow(0,0)}\frac{x·y}{x^2+y^2}=\lim_{r\rightarrow0}\cos(\theta)·sin(\theta)[/tex]
    So the limit doesn't exist.

    However I looked at the solution and it is differentiable. Is it possible that the solution is wrong or did I make a mistake?

    Thanks for your help.

    Edit: I don't why it doesn't TeX (at least on my browser) so:
    f(x,y) = tan(xy)*sin(1/[x^2+y^2])
    And for my attempt at a solution:
    lim(x,y->0,0) f(x,y) = lim (x,y->0,0) xy/(x^2+y^2) = lim (r -> 0) cos T*sin T
     
    Last edited: Apr 20, 2009
  2. jcsd
  3. Apr 20, 2009 #2

    tiny-tim

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    Hi springo! :smile:

    (for some reason the LateX doesn't seem to be working properly today … it's not just you! :blushing:)

    (and try using the X2 tag just above the Reply box :wink:)
    erm :redface: … sin(1/r) is not approximately 1/r, it's ≤ 1,

    so |f(x,y)| ≤ tan(xy) -> 0 :wink:
     
  4. Apr 20, 2009 #3
    Hi tiny-tim! Thanks for your answer.
    I'm not sure I understand what you meant with your answer.
    I was trying to find the limit by using polar coordinates:
    x = r·cos(t)
    y = r·sin(t)

    So lim(x,y)->(0,0) tan(x·y)·sin(1/[x2+y2])
    ~ lim(x,y)->(0,0) x·y/(x2+y2)
    ~ limr->0 r2·cos(t)·sin(t)/r2
    ~ cos(t)·sin(t)
    And so the limit doesn't exist.
     
  5. Apr 20, 2009 #4

    tiny-tim

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    Noooo … as r -> 0, sin(r) -> r but sin(1/r) does not -> 1/r, it stays ≤ 1

    try drawing it! :smile:
     
  6. Apr 20, 2009 #5
    OK, I understood! How could I make that mistake... :rolleyes:
    So how can I do this?

    Edit:
    I was thinking:
    limh->0 [f(x,y) - f(0,0) - fx(0,0)(x-0) - fy(0,0)(y-0)]/√(x2+y2)
    Since fx(0,0) = fy(0,0) = 0 (by doing the math... lol) and f(0,0) = 0... it's now:
    limh->0 f(x,y)/√(x2+yy) ~ x·y·sin(1/x2+y2)/√(x2+y2)
    And by turning into polar we have r2/r = r (the other stuff is bounded so...).
    So the limit is 0, so it's differentiable.
    Is that right?
     
    Last edited: Apr 20, 2009
  7. Apr 21, 2009 #6

    tiny-tim

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    Hi springo! :smile:
    a bit complicated … and sometimes fx(0,0) and fy(0,0) exist when, along some non-axis curve, the derivative doesn't exist.

    Just use the definition …

    what is lim [f(x,y) - f(0,0)]/r ? :smile:
     
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