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Proving cauchy criterion for limits

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the converse of the Cauchy Criterion for Limits.

    Let I be an interval that either contains the point c or has c as one of its endpoints and suppose that f is a function that is defined on I except possibly at the point c. Then the function f has limit at c iff for all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}


    2. Relevant equations



    3. The attempt at a solution

    Already proven is:

    If for all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}, then f has limit at c.

    I want to show:

    If f has limit at c, then or all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}

    Suppose the limit of f as x tends to c is L.

    Since f has limit at c, then or all ε>0 there exists a δ>0 such that |f(x)-L|<ε for all x,y in I that satisfy |x-c|<δ

    Then |f(x)-f(y)+f(y)-L|≤|f(y)-f(x)|+|f(y)+L|

    Since f has limit as c, let S = {t in I| 0 < |t-c| < δ}

    Choose x,y from S

    Then |f(y)-f(x)|+|f(y)+L| < |f(y)-f(x)| + ε

    This is where I am stuck. I want to show that |f(y)-f(x)| < ε.
     
  2. jcsd
  3. Oct 5, 2012 #2

    jbunniii

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    This is true but not the inequality you want. Try [itex]|f(y) - f(x)| = |f(y) - L + L - f(x)| \leq |f(y) - L| + |f(x) - L| \ldots[/itex]
     
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