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Math Help - Limit proof, absolute value problem.

  1. #1
    Junior Member
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    Limit proof, absolute value problem.

    Hi,

    I am reading a proof (in Spivak) that asserts that.

    Given 0 < |x - a| < \delta_1 and |f(x)-l|< \epsilon, and given 0 < |x - a| < \delta_2 then |f(x)-m|< \epsilon that l=m.

    We choose \delta = min(\delta_1, \delta_2) and the mid point between the two limits for epsilon. \epsilon = \frac{|l-m|}{2}.

    The proof starts thus:

    |l-m| = |l - f(x) + f(x) - m|

     \le |l - f(x)| + |f(x) - m|
     < \frac{|l-m|}{2} + \frac{|l-m|}{2}

    He continues to prove a contradiction...

    I have a problem at this point. We are substituting epsilon back in which seems simple. but does  |l - f(x)| = | f(x) - l| for all possible values of any function. Is the sign reversible even when it's a function. This seems to be required to get to the last step and I am unconvinced that it always holds. Am I missing something else?

    Thanks
    Regards
    Craig.
    Last edited by craigmain; June 2nd 2009 at 10:57 PM. Reason: clarity.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by craigmain View Post
    I have a problem at this point. We are substituting epsilon back in which seems simple. but does  |l - f(x)| = | f(x) - l| for all possible values of any function. Is the sign reversible even when it's a function. This seems to be required to get to the last step and I am unconvinced that it always holds. Am I missing something else?
    For any particular x in the domain of f f(x) is a number and so |l-f(x)|=|f(x)-l|.

    But true for any particular x in the domain of f means the same as for all x in the domain of f.

    CB
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