Essential singularity

  • Thread starter ehrenfest
  • Start date
  • #1
2,012
1

Homework Statement


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.


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
619
Why couldn't you get it? The limit doesn't even exist as you approach 0 along the positive real axis.
 
  • #3
2,012
1
Why couldn't you get it? The limit doesn't even exist as you approach 0 along the positive real axis.
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.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
619
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.
 

Related Threads on Essential singularity

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
Replies
5
Views
1K
Replies
2
Views
3K
Replies
13
Views
11K
Replies
0
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
6K
Top