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Thread: Limit proof, absolute value problem.

  1. #1
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    Limit proof, absolute value problem.

    Hi,

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

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

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

    The proof starts thus:

    $\displaystyle |l-m| = |l - f(x) + f(x) - m|$

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

    He continues to prove a contradiction...

    I have a problem at this point. We are substituting epsilon back in which seems simple. but does $\displaystyle |l - f(x)| = | f(x) - l|$ 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; Jun 2nd 2009 at 09:57 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 $\displaystyle |l - f(x)| = | f(x) - l|$ 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 $\displaystyle x$ in the domain of $\displaystyle f$ $\displaystyle f(x)$ is a number and so $\displaystyle |l-f(x)|=|f(x)-l|$.

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

    CB
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