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

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

$$f(x) = \begin{cases} \space\space\space 1 & \text{if } x \text{ is rational}\\ -1 & \text{if } x \text{ is irrrational} \end{cases}$$

I proceed as follows:

\begin{align} \\ & \text{Let } \forall \epsilon > 0 \\ & \text{Choose } \delta = \min\{1, \dfrac{\epsilon}{2}\} \\ & \text{Assume } 0 < |x| < \delta \end{align}

I'm fairly certain that the limit fails to exist because the one-sided limits are not equal to each other, but I'm not certain of how to proceed in proving that the limit is non-existent.

• Use the density of $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ to show that $f$ can take the values $1$ and $-1$ on each neighbourhood of zero. Commented Feb 17, 2015 at 11:17
• @Nicolas I don't quite follow. Commented Feb 17, 2015 at 11:19

Let us suppose a number $z$ exists with $$\lim_{x\to 0} f(x) = z$$

Given a $\epsilon$ challenge, we need to come up with a positive $\delta$ such that $$\left\lvert f(x) - z \right\rvert < \epsilon$$ for all $x \in U = (-\delta, 0) \cup (0, \delta)$.

Now despite the choice of $\delta$, each $U$ will always contain both rational and irrational numbers. That means we would need a $z$ with both $$\left\lvert 1 - z \right\rvert < \epsilon \wedge \left\lvert -1 - z \right\rvert < \epsilon$$ There is no such $z$, if $\epsilon \le 1$: $$\left\lvert 1 - z \right\rvert < 1 \Rightarrow z \in I = (0, 2) \\ \left\lvert -1 - z \right\rvert \Rightarrow z \in J = (-2, 0)$$ but $I \cap J = \emptyset$.

Tip: Construct two sequences $(x_n)$, $(y_n)$ with $x_n, y_n \to 0$ but $f(x_n) = 1$, $f(y_n) = -1$ for all $n \in \mathbb N$.

Try to construct two sequences $(u_n)$ and $(v_n)$ where :

$u_n \to 0, v_n \to 0$ and $\lim f(u_n) \neq \lim f(v_n)$

Suppose that $f(x)\rightarrow\ell$ when $x\rightarrow0$, then for all $0<\varepsilon<1$ there is a $\delta>0$ such that $$\left|f\left(x\right)-\ell\right|<\varepsilon$$ on $\left[0,\delta\right]$ (same for non-positive values). But by density, you have at least one $q\in\mathbb{Q}$ and one $r\in\mathbb{R}\setminus\mathbb{Q}$ in your interval $\left[0,\delta\right]$ : since $f\left(q\right)=1$ , we must have $\ell=1\pm\varepsilon$ , and also, since $f\left(r\right)=-1$ we must have $\ell=-1\pm\varepsilon$ , that is impossible for $\varepsilon$ small enough.

To prove that function $f$ does not have a limit at $x$, you need to establish that there is an $\epsilon >0$ such that for all $\delta >0$, there is an $y$ such that $|x-y| < \delta$ and $|f(x) - f(y)| > \epsilon$.

For your function, the "jump" in values is from $1$ to $-1$, so you need to choose an $\epsilon < | 1 - (-1)| = 2$, so pick $\epsilon = 1$ lets say. Now, you need to come up with a $y$ for every $\delta$ such that $|x - y|<\delta$ and so that the value of $|f(0) - f(y)| > 1$. To make that happen, you need y to be irrational, and "close" to $0$. Try rational multiples of an irrational, and you should be able to work it out.