- #1

- 52

- 0

[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.