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Homework Help: Epsilon Delta Proof, need clarification

  1. Sep 26, 2004 #1
    Prove that

    lim_{x \rightarrow c} \ \ \frac{1}{x}=\frac{1}{c} \ ,c\neq0


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


    0<|x-c|<\delta \ \Rightarrow \ | \ \frac{1}{x}=\frac{1}{c}|<\epsilon



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

    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:


    |c|=|c-x+x| \leq |c-x|+|x|



    |x| \geq |c|-|x-c|

    Thus if we choose


    \delta \leq \frac{|c|}{2}


    then we can succeed in making
    |x| \geq \frac{|c|}{2}

    Finally, if we also require


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

    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.
  2. jcsd
  3. Sep 26, 2004 #2


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    Dearly Missed

    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.
  4. Sep 26, 2004 #3
    Some more thoughts...

    1. As Arnildo wrote, as c!=0, there is an interval
    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.
  5. Sep 27, 2004 #4
    Thank you very much guys
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