Date: Jan 26, 2013 5:56 PM
Author: Charles Hottel
Subject: Limit Problem

I am having a problem following an example in my book.
I understand the concept of limit but sometimes I get confused
manipulating expressions with absolute values in them. Here is the problem:

Prove lim(x->c) 1/x = 1/c, c not equal zero

So 0 < | x-c| < delta, implies |1/x - 1/c| < epsilon

|1/x - 1/c| = | (c-x) / {xc}| = 1/|x| * 1/|c| * (x-c) < epsilon

Factor 1/|x| is troublesome if x is near zero, so we bound it to keep it
away from zero.

So |c| = |c - x + x| <= |c-x| + |x| and this imples |x| >= |c| - |x-c|

I think I understand everything up to this point, but not the next steps,
which are

If we choose delta <= |c|/2 we succeed in making |x| >= |c| / 2.
Finally if we require delta <= [(epsilon) * (c**2)} / 2 then

[1/|x| * 1/|c| * |x-c|] < [1 / (|c|/2)] * [1/|c|] * [((epsilon) *
(c**2)) / 2] = epsilon

How did they know to choose delta <= |c|/2?

How does that lead to |x| > |c|/2 implies 1/|x| < 1/(|c|/2) ?

I did not sleep well last night and I feel I must be missing something
that would be obvious if my head was clearer. Thanks for any help.