- #1
haribol
- 52
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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.
[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.