Let $$ f:[0,1]\to [0,1]\\ f(x)= \begin{cases} x, & x\notin\Bbb Q\\ 0, & x\in \Bbb Q \end{cases} $$

Prove the $\lim\limits_{x\to x_0}f(x)$ does not exist for $x_0 \in (0,1]$.


I don't know how to do this using the $\varepsilon-\delta$ definition, I tried supposing $\lim f(x)= \ell$ and going for a contradiction, but I couldn't reach any, I separated in the cases $\ell=0, \ell\neq 0$, but I couldn't get far with the $\ell\neq 0$ .

Could someone give me some hints/instructions on how to prove this? I'd like to solve it myself (at least as much as I can), rather than getting a complete solution.

  • $\begingroup$ Starting point is to work with $|l-x_0| $ because $0 \leq l \leq 1$ $\endgroup$ – DeepSea Feb 22 '16 at 4:10
  • $\begingroup$ @Kf-Sansoo Why can you say that $ 0\leq \ell\leq 1$? $\endgroup$ – YoTengoUnLCD Feb 22 '16 at 4:17

HINT: If $L \neq 0$ try to take rational number between $x_0$ and $x_0+\delta$ and $\epsilon = \frac{|L|}{2}$

If $L=0 $ try to take irrational number between $x_0$ and $x_0+\delta$ and $\epsilon = x_0$

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  • $\begingroup$ Thanks! I think I got it, I'll edit in my work in a moment. Could you give me a hint with the $\ell=0$ case? I thought I had gotten it, but I was wrong... $\endgroup$ – YoTengoUnLCD Feb 22 '16 at 4:40
  • $\begingroup$ if $l = 0$ let $0 < epsilon < x_0$ $\endgroup$ – fleablood Feb 22 '16 at 7:26

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