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Thread: [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 $\displaystyle f(x)=\left\{\begin{array}{cc}0,&\mbox{ if x is irrational}
    \\1, & \mbox{ if x is rational} \end{array}\right.$. Show that $\displaystyle \lim \limits_{x \rightarrow a} f(x)$ does not exist for any a

    *EDIT* I deleted the second problem so I could posted without double-posting
    Last edited by akolman; Mar 14th 2008 at 12:18 AM.
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  2. #2
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    #1. First say $\displaystyle L>0$. Suppose that $\displaystyle \lim_{x\to a}f(x) = L$ then $\displaystyle 0<|x-a|<\delta \implies |f(x) - L| < L/2$ (by picking $\displaystyle \epsilon = L/2>0$). But since the is a rational $\displaystyle x_0$ with $\displaystyle 0<|x_0-a|<\delta$ for no matter $\displaystyle \delta>0$ it would mean that $\displaystyle f(x_0)=0$ and so $\displaystyle |f(x_0)-L| = L < L/2$ this is a contradiction. If $\displaystyle L<0$ then choose $\displaystyle \epsilon = -L/2 > 0$ and apply same argument. If $\displaystyle L=0$ then it means $\displaystyle |f(x)| < \epsilon$ for all $\displaystyle 0<|x-a|<\delta$. Take then $\displaystyle \epsilon = 1$ and it would mean $\displaystyle |f(x)|<1$ for all $\displaystyle 0<|x-a|<\delta$, but since there is irrational $\displaystyle x_0$ then it would mean $\displaystyle |f(x_0)| = 1 < 1$ which is a contradiction.
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