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Thread: Precise definition of a limit

  1. #1
    Member Chokfull's Avatar
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    Precise definition of a limit

    I need help with this problem...
    prove that the limit

    $\displaystyle lim x-->0 f(x)$
    such that $\displaystyle f(x) = 0$ when $\displaystyle x$ is rational; $\displaystyle 1$ when $\displaystyle x$ is irrational

    does not exist. Thanks to anyone who helps!

    yes, I think I should learn LaTex...
    Last edited by Chokfull; Jul 3rd 2009 at 10:39 AM.
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  2. #2
    MHF Contributor

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    You have some real issues with notation.
    What you have posted is almost unreadable.
    If you want to receive real help here then you should learn LaTex.


    Your function is $\displaystyle
    f(x) = \left\{ {\begin{array}{*{20}c}
    {0,} & {x \in \mathbb{Q}} \\
    {1,} & {else} \\

    \end{array} } \right.$.

    Now you need to recall that any neighborhood of 0 contains infinitely many of both rational and irrational numbers.
    There has got to a contradiction there if that limit exists.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    This is easy with sequences.

    Show that there is a sequence $\displaystyle (x_i)_{i\in \mathbb{N}}$ of irrational points converging to 0; then if $\displaystyle f$ is continuous at 0 we must have $\displaystyle \lim f(x_i) = f(0) = 0$; but $\displaystyle \lim f(x_i) = \lim 1 = 1$ hence $\displaystyle f$ is not continuous at 0 (or anywhere for that matter).

    As a side note, a funny thing is that the function

    $\displaystyle g(x) = \left\{ {\begin{array}{*{20}c}
    {0,} & {x \in \mathbb{Q}} \\
    {x,} & {else} \\

    \end{array} } \right.$

    is continuous at 0.

    I also second Plato's advice to learn LaTeX. (At least you're trying a bit!)
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  4. #4
    MHF Contributor matheagle's Avatar
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    Both sets are dense.
    So any interval about zero, no matter how small, will always contain rationals and irrational numbers.
    Thus for all x in $\displaystyle (-\epsilon, \epsilon)$ f(x) will be both 0 and 1.
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