Prove that if $f(x)=x$ for rational $x$, and $f(x)=-x$ for irrational $x$, then $\lim\limits_{x \to a} f(x)$ does not exist if $a \neq 0$.
The solution to this problem is intuitively clear, and the solution from the solution manual provides one such descriptive solution. I'd like to know what a more rigorous solution would look like.
My complete solution considers four cases for $a \neq 0$, though I only present one below. The other three are analogous. There is also the case where $a=0$.
$$f(x)=\begin{cases} x, \text{ } x \in \mathbb{Q} \\ -x, \text{ } x \in \mathbb{Q'} \end{cases}$$
$$x \in \mathbb{Q} \implies |f(x)-l|=|x-l|$$ $$x \in \mathbb{Q'} \implies |f(x)-l|=|-x-l|=|-(x+l)|=|x+l|$$
Case 1: $a>0$, $l>0$
If $x \in \mathbb{R}:a-\delta<x<a+\delta$, $x \in \mathbb{Q'}$ then $|f(x)-l|=|x+l|$
Let $x_+ \in \mathbb{Q'}:0<a<x_+<a+\delta$.
Then $|f(x_+)-l|=|x_++l|>|a+l| \geq |a|$
Therefore, we have shown that
$$a>0,l>0, \forall \epsilon>0: 0<\epsilon<a, |x-a|<\delta \implies\exists x:|f(x)-l|>a>\epsilon$$
$\implies \lim\limits_{x\to a}f(x)=l$ is false $\forall l$.
Case 2: $a>0$, $l<0$ This proof is analogous to Case 1's proof, but we look at a $x_+ \in \mathbb{Q}$ instead of $\mathbb{Q'}$.
Case 3: $a<0$, $l>0$ Analogous proof, but we look at an $x_- \in \mathbb{Q}:a-\delta<x_-<a$
Case 4: $a<0$, $l<0$ Analogous proof, but we look at an $x_- \in \mathbb{Q'}: a-\delta<x_-<a$
Case 5: $a=0$, $l=0$
Let $|x|<\epsilon$, $\epsilon>0$.
$$x \in \mathbb{Q} \implies |f(x)-l|=|x|<\epsilon$$ $$x \in \mathbb{Q'} \implies |f(x)-l|=|x|<\epsilon$$
Therefore
$$\forall \epsilon>0, |x|<\epsilon \implies |f(x)|<\epsilon$$
$$\implies \lim\limits_{x \to 0} f(x)=0$$