How do you prove e^{1/z} has an essential singularity at z=0?

[ehrenfest]

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
$$\lim_{z\to 0} z^k e^{1/z}$$
does not exist for any natural number k, but I couldn't get it.

Homework Equations

The Attempt at a Solution

 

[Dick]

Why couldn't you get it? The limit doesn't even exist as you approach 0 along the positive real axis.

 

[ehrenfest]

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

$$\lim_{x\to 0} \lim_{k\to \infty}\sum _{n=0}^k \frac{x^{-n+k}}{k!}$$

But the double limit makes that especially hard to evaluate.

 

[Dick]

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.