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Beginner's Epsilon Delta Proof: Help needed

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data This is my first delt/epsilon proof ever, so please understand if I seem ignorant.

    e=epsilon
    d = delta

    Let f(x) = 1/x for x>0

    If e is any positive quantity, find a positive number d, which is such that:

    if 0 < |x-2| < d, then |f(x) - 1/2| < e


    2. Relevant equations
    I don't really know of any :s


    3. The attempt at a solution

    |1/x - 1/2| < e
    |2/x - 1| < 2e
    |x/2 - 1| > 1/2e
    |x - 2| > 1/e

    and |x-2| < d

    Therefore, 1/e < d

    Is this sufficient? It says find a positive d, and I've only come up with an inequality with respect to e. Again, this is my first ever d/e proofs, so if I've overlooked some tremendously obvious error, I'm sorry.
     
  2. jcsd
  3. Sep 24, 2011 #2

    LCKurtz

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    It's OK to start with an exploratory argument like this, but how did you get from the second step to the third step?
     
  4. Sep 24, 2011 #3
    I took the reciprocal of both sides - is that allowed?
     
  5. Sep 24, 2011 #4

    LCKurtz

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    No law against it. Not sure why you want the reciprocal but, given that's what you want to do, do you think the reciprocal of
    [tex]\frac 2 x -1\hbox{ is }\frac x 2 -1\hbox{?}[/tex]
     
  6. Sep 24, 2011 #5
    Oh, I see my mistake. Thank you.

    I don't really know: I took the reciprocal to try to get d and e in the same form, so I may choose d.

    Could I have a hint to a more efficient method?
     
  7. Sep 24, 2011 #6

    LCKurtz

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    Your exploratory start was OK:

    [tex]\left|\frac 1 x - \frac 1 2\right| < \epsilon[/tex]

    Start by simplifying that left side by combining the fractions. The idea is to see how close x needs to be to 2 to make the inequality work.
     
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