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Math Help - Epsylons and deltas

  1. #1
    Junior Member shannu82's Avatar
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    Epsylons and deltas

    let f(x) = 1/x for x>0. If E is any positive quantity (i.e. E >0), find a positive number S (i.e. S>0) which is such that :
    if 0<|x-3|<S, then |f(x) - 1/3| < E

    Note: E - epsylon, and S - delta

    Am not sure of how to start. Any help would be appreciated.
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  2. #2
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    Solve \left|f(x) - \frac{1}{3}\right| < \epsilon for |x - 3|.


    \left|f(x) - \frac{1}{3}\right| < \epsilon

    \left|\frac{1}{x} - \frac{1}{3}\right| < \epsilon

    \left|\frac{3}{3x} - \frac{x}{3x}\right| < \epsilon

    \left|\frac{3 - x}{3x}\right| < \epsilon

    \frac{|3 - x|}{|3x|} < \epsilon

    \frac{1}{3|x|}|x-3| < \epsilon

    \frac{1}{|x|}|x - 3| < 3\epsilon.


    Define M such that \frac{1}{|x|} \leq M, then

    |x - 3| < \frac{3\epsilon}{M}.

    So we would be able to choose \delta = \frac{3\epsilon}{M}.


    There is no M such that \frac{1}{|x|} < M for all x, but we can restrict the value of x.


    If we restrict |x - 3| < 1 (i.e. so that \delta = 1)

    then -1 < x - 3 < 1

    2 < x < 4

    \frac{1}{4} < \frac{1}{x} < \frac{1}{2}.


    Thus \frac{1}{|x|} < \frac{1}{2}.


    So set M = \frac{1}{2} so that we can define \delta to be

    \delta = \min\left\{1, \frac{3\epsilon}{\frac{1}{2}}\right\}

    \delta = \min\{1, 6\epsilon\}.


    Now all that's left is to write the proof.

    Proof: Let \epsilon > 0 and define \delta = \min\{1, 6\epsilon\}. Then if 0 < |x - 3| < \delta, we have

    \left|\frac{1}{x} - \frac{1}{3}\right| = \left|\frac{3}{3x} - \frac{x}{3x}\right|

     = \left|\frac{3 - x}{3x}\right|

     = \frac{|x - 3|}{3|x|}

     < \frac{1}{2}\left(\frac{|x - 3|}{3}\right) since \frac{1}{|x|} < \frac{1}{2}

     < \frac{1}{2}\left(\frac{6\epsilon}{3}\right) since |x - 3| < \delta and \delta < 6\epsilon

     = \epsilon.


    Thus 0 < |x - 3| < \delta \implies \left|f(x) - \frac{1}{3}\right| < \epsilon.
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