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Epsilon Delta Proof, need clarification

  • Thread starter haribol
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  • #1
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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.
 

Answers and Replies

  • #2
arildno
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1. Since c doesn't equal 0, there exists a (tiny) region about c which doesn't contain 0. This region is what we care about.
2. The negative of the first difference has equal magnitude as the difference itself.
 
  • #3
1,569
1
Some more thoughts...

1. As Arnildo wrote, as c!=0, there is an interval
I:=[c-z,c+z]
that doesn't contain the origin. If you like, z can be half the distance between c and 0 to ensure
a. if c<0 that c+z<0
b. if c>0 that c-z>0.

z could also be 3/4 the distance between 0 and c or some such. A picture might be helpful.

On I, 1/|x| can be "bounded" which means that we can find a number M such that for x in I, 1/|x| <= M. You can find a formula for M depending on c and z but NOT x. This M is used in step 8 somewhere.

You have |1/x - 1/c| < (M/|c|) |x-c| < (M/|c|) delta.

Then if delta = epsilon * (|c|/M), then we'd have that for an epsilon>0 there is a delta such that if |x-c|<delta, then |1/x-1/c|<epsilon.
 
  • #4
48
0
Thank you very much guys
 

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