Recent content by Sartre

Hallsofivy: first of all, thanks for the quick response! I am very grateful for this :smile: on problem #1; isn't the taylor series e^{z} for \frac{z^{n}}{n!}? then you get z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + ... + \frac{z^{n}}{n!} so if you divide by z you get 1 + z + \frac{z^{2}}{2!}...

Homework Statement 1) \frac{e^{z}-1}{z} Locate the isolated singularity of the function and tell what kind of singularity it is. 2) \frac{1}{1 - cos(z)} z_0 = 0 find the laurant series for the given function about the indicated point. Also, give the residue of the function at the...

Actually I reviewed this part of calculus a couple of days ago. So it is my bad om that one ;) And the simplest proof of seeing that the limit doesn't exist is graphical.

I don't think you should just say that the limit is something or other. If you want to take limits in simple non-series functions, then you should look at the outer rims of the function. For example sin(x) is defined as -1 < x < 1. Now you must work out if the limit is negative or positive...