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Multiple limits

  1. Sep 22, 2009 #1

    disregardthat

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    What conditions must f satisfy if

    [tex]\lim_{x \to a} \lim_{y \to b} f(x,y)=\lim_{y \to b} \lim_{x \to a} f(x,y)[/tex]

    where [tex]\lim_{x \to a} f(x,y)[/tex] and [tex]\lim_{y \to b} f(x,y)[/tex] exists and are finite?
     
    Last edited: Sep 23, 2009
  2. jcsd
  3. Sep 23, 2009 #2

    disregardthat

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  4. Sep 29, 2009 #3
    Without condition you're statement is not true, consider
    [tex] \frac{y^2}{x^2+y^2}. [/tex]
     
  5. Sep 29, 2009 #4

    uart

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    Good question. I'm fairly sure that continuity in both x and y would be a sufficient condition.
     
  6. Sep 29, 2009 #5
    f should be continuous at (a,b) ... that is, continuous as a function of two variables. Continuous separately in each of x and y is not enough.
     
  7. Sep 29, 2009 #6

    uart

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    Arh yes, thanks for the clarification g_edgar. :)

    BTW, what is the simplist definition of continuity in this case. I was thinking of something like :

    [itex] \exists \, \, \epsilon > 0 \, : \, |f(x+dx,y+dy) \, - \, f(x,y) | \leq \, k \, ||(dx,dy)|| [/tex] whenever [itex] ||(dx,dy)|| \leq \epsilon [/itex].

    Is that basically correct?
     
  8. Sep 30, 2009 #7
    If [tex] f [/tex] is continuous the we certainly have [tex] \lim_{x\rightarrow a} \lim_{y\rightarrow b} f(x,y) = \lim_{y\rightarrow b} \lim_{x\rightarrow a} f(x,y). [/tex]
    But you do not need that much. Consider
    [tex] g(x,y) = \frac{xy}{x^2+y^2}. [/tex]
    The function is discontinuous at [tex](0,0)[/tex], since [tex] \lim_{t\rightarrow 0} g(t,t) = 1/2 \neq 0 = \lim_{t\rightarrow 0} g(t,0) [/tex].
    But we have [tex] \lim_{x\rightarrow 0} \lim_{y\rightarrow 0} g(x,y) = \lim_{y\rightarrow 0} \lim_{x\rightarrow 0} g(x,y)=0, [/tex] and [tex]\lim_{x\rightarrow 0} g(x,y) = 0 = \lim_{y\rightarrow 0} g(x,y)[/tex] exist and are finite.
     
  9. Sep 30, 2009 #8
    No that's not it. Try again!
     
  10. Oct 3, 2009 #9

    disregardthat

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    Thanks for the replies.

    Now, consider [tex]f(x,y)=x^y[/tex] on [tex](0,\infty)[/tex] in both variables. Is the function is continous in both variables on the interval, and not only seperately? We have existing limits as [tex]x[/tex] and [tex]y \to 0[/tex] independently. They are 0 and 1 respectively. However, the resulting limit depends on the order of the limit composition.

    How do you account for this example? What conditions do f fail to satisfy? And what is the difference between continuity in two variables, and continuity in two variables seperately?
     
  11. Oct 3, 2009 #10
    Yes, [itex]f[/itex] is continuous on the whole product set [itex](0,\infty) \times (0,\infty)[/itex].

    The condition that fails: [itex]f[/itex] is not continuous at the point [itex](0,0)[/itex]

    Remember what I said back there? "f should be continuous at (a,b)" ?
     
  12. Oct 3, 2009 #11

    disregardthat

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    How is continuity defined in a point for a function of 2 or more variables?
     
  13. Oct 3, 2009 #12
    A function f is continuous at a point v, if for every e>0 there is a d>0 such that
    ||f(v + u) - f(v)|| < e whenever ||u|| < d.
     
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