@Knight Since you won't give up on this, suppose it did. Then there would exist $\delta>0$ such that if $|x-a|<\delta$ then $|f(x) - \ell|<|a|/2$. Since the set of such points $x$ is an interval around $a$, there is a rational such $x$, say $x_0$, and an irrational one, say $x_1$, such that $x_1$ is further from $0$ than $a$, i.e. $|x_1|>|a|$. We get the inequalities $|f(x_0) - \ell | = |\ell | < |a|/2$ and $| f(x_1) - \ell | = |x_1 - \ell | < |a|/2$. Then
$$ |a|<|x_1| \le |\ell| + |x_1 - \ell | < |a|$$

Suppose that $\lim\limits_{x\to a}f(x)=L$. Then the definition of a limit says
$$
\forall\epsilon\gt0,\exists\delta\gt0:\forall x,|x-a|\le\delta\implies|f(x)-L|\le\epsilon\tag1
$$
Choose $\epsilon=a/4$. For the $\delta\gt0$ guaranteed above, we have that
$$
|x-a|\le\delta\implies|f(x)-L|\le a/4\tag2
$$
we can pick $x_1\in\mathbb{Q}$ so that $|x_1-a|\le\min(\delta,|a|/4)$ and $x_2\not\in\mathbb{Q}$ so that $|x_2-a|\le\min(\delta,|a|/4)$. Note that since $|a|\le|x_1|+|x_1-a|$ and $|x_1-a|\le|a|/4$, we have