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Essential singularity

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data
    http://en.wikipedia.org/wiki/Essential_singularity

    What is the best way to prove that e^{1/z} has an essential singularity at z=0? I have tried showing that
    [tex]\lim_{z\to 0} z^k e^{1/z}[/tex]
    does not exist for any natural number k, but I couldn't get it.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 31, 2008 #2

    Dick

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    Why couldn't you get it? The limit doesn't even exist as you approach 0 along the positive real axis.
     
  4. Mar 31, 2008 #3
    How do you prove that? I tried using the definition of e^x

    [tex]\lim_{x\to 0} \lim_{k\to \infty}\sum _{n=0}^k \frac{x^{-n+k}}{k!}[/tex]

    But the double limit makes that especially hard to evaluate.
     
  5. Mar 31, 2008 #4

    Dick

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    It's easiest if you take x=1/z for z real. Then the limit becomes lim(x->infinity) e^x/x^k. Now 'everybody knows' e^x approaches infinity faster than any power of x. But if you want to show it, use l'Hopital k times.
     
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