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Math Help - [SOLVED] two (tough?) limit proofs

  1. #1
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    [SOLVED] two (tough?) limit proofs

    Hi, I'm having trouble to write a formal proof for the following.

    1. Let f(x)=\left\{\begin{array}{cc}0,&\mbox{ if x is irrational}<br />
\\1, & \mbox{ if x is rational} \end{array}\right.. Show that \lim \limits_{x \rightarrow a} f(x) does not exist for any a

    *EDIT* I deleted the second problem so I could posted without double-posting
    Last edited by akolman; March 14th 2008 at 01:18 AM.
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  2. #2
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    #1. First say L>0. Suppose that \lim_{x\to a}f(x) = L then 0<|x-a|<\delta \implies |f(x) - L| <  L/2 (by picking \epsilon = L/2>0). But since the is a rational x_0 with 0<|x_0-a|<\delta for no matter \delta>0 it would mean that f(x_0)=0 and so |f(x_0)-L| = L < L/2 this is a contradiction. If L<0 then choose \epsilon = -L/2 > 0 and apply same argument. If L=0 then it means |f(x)| < \epsilon for all 0<|x-a|<\delta. Take then \epsilon = 1 and it would mean |f(x)|<1 for all 0<|x-a|<\delta, but since there is irrational x_0 then it would mean |f(x_0)| = 1 < 1 which is a contradiction.
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