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Math Help - 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

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

    does not exist. Thanks to anyone who helps!

    yes, I think I should learn LaTex...
    Last edited by Chokfull; July 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 <br />
f(x) = \left\{ {\begin{array}{*{20}c}<br />
   {0,} & {x \in \mathbb{Q}}  \\<br />
   {1,} & {else}  \\<br /> <br />
 \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 (x_i)_{i\in \mathbb{N}} of irrational points converging to 0; then if f is continuous at 0 we must have \lim f(x_i) = f(0) = 0; but \lim f(x_i) = \lim 1 = 1 hence f is not continuous at 0 (or anywhere for that matter).

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

    g(x) = \left\{ {\begin{array}{*{20}c}<br />
   {0,} & {x \in \mathbb{Q}}  \\<br />
   {x,} & {else}  \\<br /> <br />
 \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 (-\epsilon, \epsilon) f(x) will be both 0 and 1.
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