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Math Help - Epsilon Delta Proof

  1. #1
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    Epsilon Delta Proof

    When proving that limit (x->5) 1/x = 0.2, find the largestδ > 0 that works for ε = 0.0025 = 1/400

    My attempt: | 1/x - 1/5 | = | (5-x)/5x | = | (x-5) /5x | <
    ε. | (x-5) /5x | < 1/400 whenever 0 < | x-5| < δ. So for some real number M, if 1/|5x| < M, then | (x-5) /5x | < M |x-5|. Hence | x-5| < ε/M = δ.

    Now |x-5| < 1 implies 4<x<6 which further implies 1/30 < 1/|5x| < 1/20, thus M= 1/20 is suitable. So we can take
    δ = min{1,20*ε}. That is the largest δ is 20/400 = 0.05.

    Is this argument correct?
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  2. #2
    Junior Member Nehushtan's Avatar
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    Re: Epsilon-delta proof

    Work backwards.

    -\frac1{400}<\frac1x-\frac15<\frac1{400} \iff \frac{79}{400}<\frac1x<\frac{81}{400} \iff \frac{400}{81}<x<\frac{400}{79} \iff -\frac5{81}<x-5<\frac5{79}

    So we want the maximum \delta such that -\delta<x-5<\delta\ \Rightarrow\ -\frac5{81}<x-5<\frac5{79}. This is clearly \delta=\frac5{81}.
    Last edited by Nehushtan; March 2nd 2013 at 04:17 AM.
    Thanks from pessimist92 and topsquark
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  3. #3
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    Re: Epsilon Delta Proof

    Hello, pessimist92. Your procedure is correct if you want to construct an epsilon-delta proof. It gives you a formula so that for any given epsilon, you can construct a delta that works. But the question asked for the *largest* delta that works for this particular epsilon, so you need to do what Nehushtan says. Your formula gives \delta=\frac{1}{20}, which is less than \frac{5}{81}.

    - Hollywood
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  4. #4
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    Re: Epsilon Delta Proof

    Thanks to both of you. Now I understand the logic of these type of Epsilon-delta questions.
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