Show that $\lim_{x\to0}f(x)$ doesn't exist using $\epsilon$-$\delta$

Question

$$f(x) = \begin{cases}0\,;\,x\in\mathbb Q\\ 1\,;\,x\in\mathbb Q^c\end{cases}$$ Show that $\lim\limits_{x\to0}f(x)$ doesn't exist using epsilon delta definition. How would I separate rational and irrational numbers if the definition deals with open intervals?

Answer 1

Hint.

Open intervals around $0$ are of the form $(-\delta, \delta)$. What can you say about rationals and irrationals in those intervals? What might the limit be if it were to exist?

Answer 2

$$|f(\text{rational})-f(\text{irrational})|>\frac12$$

EDIT (Answer to the comment)

Take $\epsilon=\frac12$. For every $\delta>0$ there exists an irrational number $t$ in $(-\delta,\delta)$. Then $$|f(t)-f(0)|>\epsilon$$

We have shown the negation of the statement "$f$ is continuous at $0$".

Answer 3

the limit does not exist since this holds. $\forall \delta \in \mathbf{Q}: f(\delta)=0 \land f(\delta\frac{(1+\pi)}{5})=1$ which means that for any $\delta$ in the rationals there is a smaller irrational and so the function near 0 just oscillates between 0 and 1

EDIT. choose an arbitrary rational $\delta$ arbitrarily close to 0

let's say $\delta=0.01$

now if we multiply $\delta$ by $\frac{(1+\pi)}{5}$ it becomes $0.01\frac{(1+\pi)}{5}=0.00828...$ which an irrational smaller (in absolute sense) than the $\delta$ we first chose

Evaluating your function in both these numbers gives for $\delta$ the value 0 and for $\delta\frac{(1+\pi)}{5}$ gives the value 1.

this argument works for any rational $\delta$ we pick, so near zero we have demonstrated that the function oscillates.

Answer 4

We are to prove that, for every $l \in \mathbb{R}$ there is some $\varepsilon > 0$ such that for every $\delta > 0$ there is some $0 < |x| < \delta$ such that $|f(x) - l| \geq \varepsilon$. (You sufficiently familiar with such a statement?)

Let $l \in \mathbb{R}$. If $l \geq 1$, then for all $x \in \mathbb{Q}$ we have $|f(x) - l| = l$; taking $\varepsilon := l/2$ suffices. If $0 < l < 1$, let $m := \min \{ l, 1-l \}$. Then for all $x \in \mathbb{R}$ we have $|f(x) - l| \geq m$; taking $\varepsilon := m/2$ suffices. If $l \leq 0$, then for all $x \in \mathbb{Q}^{c}$ we have $|f(x) - l| = 1 + |l|$; taking $\varepsilon := (1+|l|)/2$ suffices.