Epsilon Delta Proof, need clarification

Prove that

[tex]
lim_{x \rightarrow c} \ \ \frac{1}{x}=\frac{1}{c} \ ,c\neq0
[/tex]

Proof

We must find [tex]\delta[/tex] such that:

1.

[tex]
0<|x-c|<\delta \ \Rightarrow \ | \ \frac{1}{x}=\frac{1}{c}|<\epsilon
[/tex]

Now,

2.

[tex]| \ \frac{1}{x}=\frac{1}{c}|=| \ \frac{c-x}{xc}|= \frac{1}{|x|} \cdot \frac{1}{|c|} \cdot |x-c|
[/tex]

The factor [tex]\frac{1}{|x|}[/tex] is not good if its near 0. We can bound this factor if x can be away from 0. Note:

3.

[tex]
|c|=|c-x+x| \leq |c-x|+|x|
[/tex]

so

4.

[tex]
|x| \geq |c|-|x-c|
[/tex]

Thus if we choose

5.

[tex]
\delta \leq \frac{|c|}{2}
[/tex]

6.

then we can succeed in making
[tex]
|x| \geq \frac{|c|}{2}
[/tex]

Finally, if we also require

7.

[tex]\delta \leq \frac{\epsilon c^2}{2}[/tex]

then,

8.

[tex]
\frac {1}{|x|} \cdot \frac{1}{|c|} \cdot |x-c| < \frac {1}{|c|} \cdot \frac {1}{|c|/2} \cdot \frac{\epsilon c^2}{2} = \epsilon
[/tex]


My questions

1. On step 2, its said to "bound" the factor 1/|x|. What does this mean?

2.How did the |c-x| on step 3 went to |x-c| on step 4?

3. How did they choose [tex]\delta[/tex] to be that value?

Any help on this subject is very much appreciated, thank you in advance.

This example is from "Calculus 8th Edition" by Varberg, Purcell and Rigdon.

 

Thank you very much guys