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Iterated limit? double limit?

  1. Sep 13, 2005 #1
    I am confused about things about "iterated limit" and "double limit"
    I am doing something about complex variables, and learning to derive the Cauchy-Riemann Equation.
    The concept of "iterated limit" to me is that, supossing 2 independt variables x&y, it is taking the limit either x first then y or y first then x. Geometrically, it is representing two paths to approach the right destination, right?

    But I can't really visualise how "double limit" really work.
    What is a double limit? I am told that it is "x and y go togather in any manner"
    What does it mean?
    I can think of some cases that x and y will approach the destination togather.
    Say, the path is y=mx as x->0. In this case, x and y will approach to 0 togather. So, is this a double limit?

    If this is, there is also a problem.
    There is a statement that "the limit, representing the derivative of a complex function, must exist as a double limit for delta z= delta x+i(delta y) approaching zero."

    I don't see why the term "double limit" is used here.

    To me, the complex derivative exists when all double limits and all iterated limits gives the same value so that the limit can be said of being independent of any paths. Am I correct?
  2. jcsd
  3. Sep 13, 2005 #2


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    You are correct. The Cauchy-Riemann equations simply impose the restriction that the limit is the same whether [itex]\Delta z[/itex] appraoches zero along the real and imaginary axis. It can be proven that this is enough to guarantee that the limit is the same along all paths in the complex plane.

    In terms of a double limit in general, that could mean something like:
    [tex]\lim_{\substack{x\rightarrow 0\\y\rightarrow 0}} x^y=1[/tex]
    [tex]\lim_{\substack{y\rightarrow 0\\x\rightarrow 0}} x^y=0[/tex]
    This really is not the same thing as delta z approaching zero, having real and imaginary parts. z is only one number, and that is how it is treated in the analytic functions of complex analysis. You could write a real number in terms of rational and irrational parts, like: [itex]x = a + b\sqrt{2}[/itex] Then you could take the limit as x approaches zero as only one number, because in real analysis we do not treat functions that perform different operations on the rational and irrational parts. Just so, in complex analysis we deal with functions that treat z as one indivisible entity.
    Last edited: Sep 13, 2005
  4. Sep 13, 2005 #3
    don't we need to specific a path for double limit?
  5. Sep 13, 2005 #4
    Don't we need to specify a path for double limit?
    like y=mx, x->0, y->0?
    And how can the two examples of double limit be calculated?
  6. Sep 13, 2005 #5

    matt grime

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    Well, the limit only exists if it is "path independent"

    let us fix that we're taking a limit as x,y both tend to 0 of some function f(x,y), let us further suppose that we want to show that limit is zero

    then we need to show that for all e>0 there is a d>0 such that for all |(x,y)|<0 that |f(x,y)<0 where | | means distance eg the euclidean distance from the origin.

    it is natrually easier to show that a limit does not exist by showing two different paths that have different limits.

    however this is a situation when we are talking about (x,y) tends to zero.

    this strictly different from a double limit
    lim x tends to 0 of lim y tends to 0 of f(x,y)
    Last edited: Sep 13, 2005
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