# Limit definition to compute some complex value function

1. Nov 29, 2013

### Genericcoder

So there is something I don't understand in the definition of limit that is applied to some problem
I have some intuition for like the rigorous limit definition but I don't have full understanding when applied to some problems.

Use definition 2 to prove lim as z → i of z^2 = -1

We must show that for given E > 0 there is a positive number x such that

|z^2 - (-1)| < E whenever 0 < |z - i| :

so we express |z^2 - (-1) | in terms of |z - i|:

z^2 - (-1) = z^2 + 1 = (z - i)(z + i) = (z - i)(z - i + 2i)

It follows from the properties of absolute value defived in Sec 1.3
that

|z^2 - (-1)| = |z - i||z - i + 2i| <= |z - i| (|z - i| + 2)

Now if |z - i| < x the right hand is less than x(x + 2) so to ensure that it is less than E, we can choose x to be smaller than either of the number E/3 and 1:

|z - i||(|z - i| + 2) < E/3(1 + 2) = 2

So Here there is alot of stuff that I don't understand like why did we need to express one value in terms of the other ? I still don't know follow what's going on here if someone could explain please because I want full understanding of these stuff. thank you.

2. Nov 29, 2013

### LCKurtz

I'm guessing you are OK to here, right? You are trying to make $z^2-(-1)$ small by getting $z$ near $i$. Now, on the right side you have $z-i$ which you can make small. That will make the right side small if the rest of it $|z-i+2i|$ isn't too big. Well, how big can it be? $|z+i|\le |z|+|i| = |z|+1$. That could presumably be very large, except for the fact we are making $z$ close to $i$. Let's say we keep $z$ within $1$ unit of $i$ so $|z-i|<1$. How big could $z$ be then? Well, $|z|-|i|\le |z-i|<1$ so $|z|<1+|i|=2$. Putting this together gives $|z+i|\le 3$ so $|z^2-(-1)|\le|z-i|\cdot 3$. That's where the overestimate of $3$ comes from. Now pick $\delta=\min\{\epsilon/3,1\}$.

Last edited: Nov 29, 2013
3. Nov 29, 2013

### Genericcoder

Oke good I understand it now thank you it makes perfect sense.