Prove that the Dirichlet function does not have a limit at $0$, with epsilon and delta

Question

I am doing a question (q-39 from ch 2.4- Stewart's Calculus) that asks me to prove that for the function

$$f(x)=\begin{cases} 0, & \text{if } x \text{ is rational} \\ 1, & \text{if } x \text{ is irrational} \end{cases}$$

the $\lim_{x\to 0}f(x)$ does not exist.

I am stuck because I have not dealt with the limits of conditional functions using delta epsilon. I understand that the start of the proof would be to assume $|f(x)-L|<\epsilon$. However the solution set suggests that it follows that $\epsilon=1/2$. How do I get $\epsilon=1/2$ from $|f(x)-L|<\epsilon$?

Answer 1

Ok, not $ε-δ$ but $\lim_{x\to0}\sup f(x)=1$, $\lim_{x\to0}\inf f(x)=0$.

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To prove it with $ε-δ$, see here:

To show that the function is not continuous at $0$, we need to find an $ε$ such that no matter how small we choose $δ$, there will be points $z$ within $δ$ of $0$ such that $f(z)$ is not within $ε$ of $f(0) = 0$. In fact, $ε=1/2$ is such an $ε$. Because the irrational numbers are dense in the reals, no matter what $δ$ we choose we can always find an irrational $z$ within $δ$ of $0$, and $f(z) = 1$ is at least $1/2$ away from $0$.

Still, quoting from the above link:

In less rigorous terms, between any two irrationals, there is a rational, and vice versa.

So $ε=1/2$ is just a choice. Any choice of $0<ε<1$ would do, but $ε=1/2$ is simple.

Answer 2

Let us argue by contradiction; let $L := \lim_{x \to 0}f(x)$ exist. If $L \leq 0$, then $|f(x) - L| > |L|$ for all irrational $x$; if $L \geq 1$, then $|f(x) - L| > |L-1|$ for all rational $x$. Hence we suppose $0 < L < 1$. Then there is some $\delta > 0$ such that $0 < |x| < \delta$ implies $$ |f(x) - L| < \min \{ L, 1-L \}/2, $$ which is impossible (why?).