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Limits with absolute values

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Let f(x, y) = |xy|.

    I want to prove that f is not continuous at (0,0).


    3. The attempt at a solution


    To prove that f is not continuous at (0,0) I think I need to show that

    [tex]\lim_{(x, y) \to (0, 0)}|xy| \neq 0[/tex]

    I'm a little confused about the |absolute value| signs and I don't know know what to do with them. Can anyone help please? (there's nothin about this in my book)
     
  2. jcsd
  3. Mar 13, 2010 #2

    tiny-tim

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    Hi math2010! :smile:
    I think it is continuous at (0,0) :confused:

    Try proving that. :smile:
     
  4. Mar 14, 2010 #3

    Well the whole problem says "Show that [tex]f \notin C^1[/tex] at (0,0)."

    In my course "[tex]f \in C^1[/tex]" means that "[tex]f_x[/tex] and [tex]f_y[/tex] exist and are continuous".

    So, I know that f(x, y) = |xy| is differentiable at (0,0). I want to prove that it does not belong to [tex]C^1[/tex], which is why I tried to show it's not continous. I'm very confused here :confused:
     
  5. Mar 14, 2010 #4

    jbunniii

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    I think the notion of [itex]C^1[/itex] is defined only if you are talking about some open set. I don't think it makes any sense to say "[itex]f \in C^1[/itex] at the point [itex](0,0)[/itex]".

    It's true that [itex]f[/itex] is both continuous and differentiable at [itex](0,0)[/itex].

    However, [itex]f[/itex] is NOT differentiable at [itex](x,0)[/itex] for [itex]x \neq 0[/itex], nor at [itex](0,y)[/itex] for [itex]y \neq 0[/itex]. Therefore if [itex]U[/itex] is an open set containing the origin, then [itex]f \not\in C^1(U)[/itex].
     
  6. Mar 14, 2010 #5

    tiny-tim

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    ah! that's not what you said originally :wink:

    So f is in C0 (the continuous functions), but not in C1 (the continuously differentiable functions)
    As jbunniii :smile: says, you need to show that f is not continuously differentiable in a neighbourhood of (0,0).

    First step … what are the derivatives of f ? :smile:
     
  7. Mar 14, 2010 #6
    Since the derivative of the absolute value function is the function [tex]\frac{x}{|x|}[/tex] (where the denominator isn't zero), I guess the derivatives of f(x,y)=|xy| are [tex]f_{x}(x,y)= f_y(x,y)= \frac{xy}{|xy|}[/tex].

    So what does this tell us? :rolleyes:
     
  8. Mar 14, 2010 #7

    tiny-tim

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    No.

    Differentiate without the ||, and then adjust it. :wink:
     
  9. Mar 14, 2010 #8
    without || it is:

    So [tex]f_x (x,y) = y[/tex]

    And I adjust it to [tex]\frac{y}{|xy|}[/tex].

    And [tex]f_y (x,y) = x[/tex] so it is [tex]\frac{x}{|xy|}[/tex]. Is this correct? :confused:
     
  10. Mar 15, 2010 #9

    tiny-tim

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    No, the magnitude will just be y, so you need to multiply by the sign of xy (or xy/|xy| or |xy|/xy).
     
  11. Mar 16, 2010 #10
    ah, you mean [tex]f_x (x,y)= \frac{y}{|y|}[/tex] and [tex]f_y(x,y)= \frac{x}{|x|}[/tex]?
     
  12. Mar 16, 2010 #11

    tiny-tim

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    But they both have magnitude 1, and they don't involve the sign of xy.
     
  13. Mar 16, 2010 #12
    What do you mean?
     
  14. Mar 17, 2010 #13
    I'm sorry, I don't really understand what you mean by saying "they don't involve the sign of xy". Do you mean it will be [tex]f_x(x,y) = y[/tex]?
     
  15. Mar 17, 2010 #14

    tiny-tim

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    If it wasn't for the ||, fx(x,y) would be y.

    The || will change the sign, but not the magnitude. :smile:
     
  16. Mar 17, 2010 #15
    So how does this enable us to show [tex]f \notin C^1[/tex]?
     
  17. Mar 17, 2010 #16

    tiny-tim

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    I think f is differentiable at (0,0), but not continuously differentiable.
     
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