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This question is from Spivak Chapter 5: Problem 20:

which says:

Prove that if $f(x) = x$ for rational $x$, and $f(x) = -x$ for irrational $x$, then $\lim_{x\to a}f(x)$ does not exist if $a \not= 0$.

My Work:

I want to base this on the previous problem which was:

Prove that if $f(x) = 0$ for irrational $x$ and $f(x) = 1$ for rational $x$, then $\lim_{x\to a}f(x)$ does not exist for any a.

Here: since the range of the values of $f(x)$ is 1, then I choose $\varepsilon$ to be $\frac{1}{4}$ and since near any a, there are both rational $x$ and irrational $x$, then for $0 < |x-a| < \delta$ if $|1 - l| < \frac{1}{4}$ then $|0 - l| > \frac{1}{4}$ and if $|0 - l| < \frac{1}{4}$ then $|1 - l| > \frac{1}{4}$ thus showing the limit does not exist.

However, for this problem, the range of the values of $f(x)$ is $2x$ yet I cannot choose $\varepsilon$ to be $\frac{x}{2}$ so it seems this same strategy would not work.

Thus how to choose $\varepsilon$ in this problem?


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I would say that this is a poor choise of $\epsilon$ which is not even dependent on the point $a$. What if $a=0.001$ ? – Belgi Oct 12 '12 at 4:52
up vote 4 down vote accepted

Hint: We are trying to prove that the function is not continuous at $a$. Let $\epsilon=|a|/2$.

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The question is to show limit does not exist so by definition isn't it to choose a $\varepsilon$ such that $|f(x) - l | >$ the chosen $\varepsilon$? – mathnoob Oct 12 '12 at 5:03
Thanks, I had a blind spot. – André Nicolas Oct 12 '12 at 5:03
thanks, got confused by the dummy variable... – mathnoob Oct 12 '12 at 13:23

An alternative approach is to use the following definition of continuity,

One can instead require that for any sequence $(x_n)_{n\in\mathbb{N}}$ of points in the domain which converges to $c$, the corresponding sequence $\left(f(x_n)\right)_{n\in \mathbb{N}}$ converges to $f(c)$. In mathematical notation, $\forall (x_n)_{n\in\mathbb{N}} \subset I:\lim_{n\to\infty} x_n=c \Rightarrow \lim_{n\to\infty} f(x_n)=f(c)\,.$

Pick up a sequence $a(n) \in Q $ with a limit point $ a \in Q' $ and apply the above definition.

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