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Can the limits of a function be imaginary?

  1. Mar 27, 2010 #1
    I was just doing some homework, and I got to thinking about this.

    So if the limit of a function is an imaginary number, does that mean that the limit does not exist? Or that it does not exist on the xy-plane, or what?

    I mean...imaginary and complex numbers exist, we just can't graph them on a Cartesian plane, right? Right?

    I've managed to confuse myself terribly.

    Please, help my brain.
  2. jcsd
  3. Mar 27, 2010 #2
    The limit of a function at a particular point can be an imaginary number...but only if the function is complex-valued.

    A function is a map from one space to another. The functions commonly encountered in introductory calculus courses are normally real-valued functions, which take a single variable out of R (the real line), and map that to another value on R. For these functions, it's impossible for a limit to be an imaginary number. (verify this! hint: can you make the function value arbitrarily close to an imaginary number?)

    Other functions, however, can take a real or complex argument and return a complex number. Now the limit of the function at a point could be imaginary, because the function value could get arbitrarily close to an imaginary number.

    So there's no great mystery about limits being complex or imaginary; if the function value can take on those values, then limits can be of those values. However, in the familiar R -> R functions that are the topic of early calculus, this just can't happen.
  4. Mar 27, 2010 #3
    Ah, thank you so much for clarifying this for me. :)

    I don't really have anyone around me that I can talk to about math, excluding my teacher, and I just had to know! Ah!

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