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Continuity of arctan

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Let zn = Arg(-1 + i/n). Find limn→∞ zn

    2. Relevant equations

    Definition of convergence of a sequence.

    3. The attempt at a solution

    Well zn = Arg(-1 + i/n) = arctan(-1/n).

    So it seems clear that limn→∞arctan(-1/n) = arctan(limn→∞ -1/n) = arctan(0) = [itex]\pi[/itex].

    Which is true if arctan is continuous.

    There's the problem, I don't know how to prove that arctan is continuous. It seems way more difficult that I expected. I can't seem to get anywhere with it.

    How do you show it?

    Or can anyone think of another way to find the limit?
     
  2. jcsd
  3. Sep 27, 2012 #2

    SammyS

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    Well, the derivative of arctan(x) is [itex]\displaystyle \frac{1}{1+x^2}\,,[/itex] which is continuous for all x.

    If arctan(x) is differentiable, what does that say about its continuity?
     
  4. Sep 28, 2012 #3
    Yea if it is differentiable then it's continuous. But I'm guessing I can't just state that the derivative of arctan is 1/(1 + x2).

    So I need to prove it.

    I've seen the derivation of it, using the triangle, but I don't think that's a legitimate proof is it?

    Wouldn't I have to use the definition of a derivative? And that seems really difficult.
     
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