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Finding the Limit of a Complex Function

  1. Oct 13, 2014 #1
    Hello everyone,

    How do I find the limit of a complex function from the definition of a limit? For instance, consider the limit

    ##lim_{z \rightarrow -3} (5z+4i)##.

    Would I simply conjecture that ##5z + 4i## approaches ##5(-3) + 4i## as ##z \rightarrow -3##; and then use the definition of a limit to justify this?
     
  2. jcsd
  3. Oct 13, 2014 #2

    Mark44

    Staff: Mentor

    Yout don't need to use the definition of the limit to find the limit value in this case. Just plug -3 in to get the limit, which is -15 + 4i. That's pretty easy; proving it will take a bit more work.
     
  4. Oct 13, 2014 #3
    I have another question. In the definition of a limit, why aren't the quantifiers reversed, so that the definition of a limit read ##\exists \epsilon >0## ##\forall \delta >0...##?
     
  5. Oct 13, 2014 #4
    [itex]\forall \epsilon > 0: \left \vert {f\left({x}\right) - L}\right \vert < \epsilon[/itex] means "no matter how small the distance between f(x) and L is", i.e., "for any degree of closeness you can think of, no matter how small".

    [itex]\exists \delta > 0: 0 < \vert x - c \vert < \delta[/itex] means that I can guarantee you there's a way to get that close. The way you get that close is by making x very close to c, but not actually at c.
     
    Last edited: Oct 13, 2014
  6. Oct 13, 2014 #5

    Mark44

    Staff: Mentor

    I've deleted my earlier post. I misread what Bashyboy wrote, mistakenly thinking that all he did was reverse the order of the quantifiers.
     
  7. Oct 14, 2014 #6
    Hmm...It still seems as though it should be reversed. It seems that, no matter how close ##z## gets to ##z_0##, ##f(z)## should also be close to ##L##; that is, for every distance ##\delta## I chose, there should exist a corresponding distance ##\epsilon##.
     
    Last edited by a moderator: Oct 14, 2014
  8. Oct 14, 2014 #7

    Mark44

    Staff: Mentor

    That's not the way to think about it. The goal is not to show that some ##\epsilon## exists - it's to show that it can be made arbitrarily small.

    Think of the whole limit business as a dialog between you (the prover) and a skeptic you're trying to convince.

    He says, "OK, can you make f(z) within 0.1 of L?"
    You say, "Sure, take ##\delta## to be 0.05. Then for every z within 0.05 of z0, f(z) is within 0.1 of L. Are you convinced?"
    He says, "Not yet. Can you get f(z) within 0.01 of L?"
    You say, "Sure, take ##\delta## to be 0.0025. Then for every z within 0.025 of z0, f(z) is within 0.01 of L. Are you convinced now?"
    .
    .
    .
    Finally, he gives up after realizing that no matter what he chooses for ##\epsilon## (this is the ##\forall \epsilon## part) you can find a ##\delta## (the ##\exists \delta## part).
     
  9. Oct 14, 2014 #8
    Okay, so I have a another complex limit I am dealing with. Here is my work:

    ##\lim\limits_{z \rightarrow (2+4i)} f(z) = 5i##, where ##f(z) = 5i##. Let ##\epsilon >0 ## be arbitrary.

    ##|f(z) - 5i| < \epsilon##

    ##|5i - 5i| < \epsilon##

    ##|5i - 5i + (z-z) + (2-2)| < \epsilon##

    ##|(5i-z-2) + (z-5i+2)| < \epsilon##

    Using the triangle inequality,

    ##|5i-z-2| + |z-5i-2| < \epsilon##

    ##|-1 \cdot (z + 2 - 5i)|| + |z-5i-2| < \epsilon##

    ##|z + 2 - 5i| + |z + 2 - 5i| < \epsilon##

    ##|z + 2 - 5i| < \frac{\epsilon}{2}##

    ##|z - 2 + 4 - 4i - i| < \frac{\epsilon}{2}##

    ##|(z - 2 - 4i) + (4-i)| < \frac{\epsilon}{2}##

    ##|z - (2 + 4i)| +|4-i| < \frac{\epsilon}{2}##

    ##|z - (2 + 4i)| < \frac{\epsilon}{2} - \sqrt{17}##

    If I let ##\delta = \frac{\epsilon}{2} - \sqrt{17}##, isn't possible that it could be negative?
     
  10. Oct 14, 2014 #9

    Mark44

    Staff: Mentor

    Since f(z) is constant, no matter what ##\epsilon## anyone chooses, you have complete freedom with ##\delta## - any value of ##\delta## will work. None of the work below is necessary.
     
  11. Oct 15, 2014 #10
    And how might one make this argument more rigorous and convincing, as I am not exactly convinced.
     
  12. Oct 15, 2014 #11

    Mark44

    Staff: Mentor

    f(z) =5i
    Prove that ##\lim_{z \to z_0}f(z) = 5i##
    Let ##\epsilon > 0## be given.
    ##|f(z) - f(z_0)| < \epsilon##
    ##\Rightarrow |5i - 5i| = 0 < \epsilon##
    Since ##|f(z) - 5i| < \epsilon## for any choice of z, take ##\delta## equal to whatever you want.

    Proofs like this one don't get any simpler than this!
     
    Last edited: Oct 15, 2014
  13. Oct 15, 2014 #12

    RUber

    User Avatar
    Homework Helper

    Be careful here. The triangle inequality states ##|a + b| \leq |a| + |b|##
    Thus, the right side might be larger in modulus than ##\epsilon##.
    Making ##|a| + |b| < \epsilon## not necessarily true.
     
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