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Epsilon-Delta Definition of Limits

  1. Oct 2, 2009 #1
    Using the epsilon-delta definition of limit, prove that the limit as x approaches 0 of f(x) equals the limit as x approaches a of f(x-a).


    I let the limit as x approaches 0 of f(x) equal A and the limit as x approaches a of f(x-a) equal B. If the absolute value of (A-B) is less than epsilon for all positive epsilon, then A=B. If A does not equal B, then let epsilon equal the absolute value of (A-B) divided by 2. Where do I go from here?
     
  2. jcsd
  3. Oct 2, 2009 #2

    LCKurtz

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    Since you are give that limit as x --> 0 of f(x) = A you have the following statement:

    Given ε >0 there is a δ > 0 such that |f(x) - A| < ε when 0 < |x| < δ

    And you are trying to prove limit as x --> a of f(x-a) = A which is:

    Given ε >0 there is a δ > 0 such that |f(x-a) - A| < ε when 0 < |x-a| < δ

    If you think about it carefully you should be able to see how the first statement implies the second one.
     
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