$f(x) =\begin{cases} x, & \text{if $x$ rational} \\ -x, & \text{if $x$ irrational} \end{cases}$, then $\lim_{x\to a}f(x)$ does not exist if $a\neq 0.$ Problem:

Prove that if $f(x) = x$ for rational $x$, and $f(x) = -x$ for irrational $x$, then $\lim_{x\to a} f (x)$ does not exist if  $a\neq 0.$

This is an exercise that appears in Spivak´s book, Calculus.
I understand the problem, however, the solution Spivak gave confused me.
Spivak's answer:

Intuitively, it is similar to what I had thought, however, there are certain parts that I don´t understand.
If $x \in (min(\frac{a}{2},a - \delta), a+\delta)$ then, certainly, $f(x) > \frac{a}{2}$ for rational $x$ and $f(x) < -\frac{a}{2}$ for irrational $x$ but my doubts arise when he says:

Since the distance between $\frac{a}{2}$ and $-\frac{a}{2}$ is $a$, this means that we cannot have $|f(x)-l| < a$ for all such $x$, no matter what $l$ is.

Shouldn´t it say that we cannot have $|f(x)-l| < \frac{a}{2}$ ? Furthermore, how can we prove that this last statement is true?
Thanks in advance.
 A: Yes, it should say that $|f(x) - l| > \dfrac{a}{2}.$
Edit
Per the comment of fleablood, immediately following this answer, what I intend is that regardless of whether $l$ is non-negative or negative, in any arbitrarily small neighborhood around $(a)$ there will exist at least one value $(x)$ such that $|f(x) - l| > \dfrac{a}{2}.$
To prove this, you have two cases:
$\underline{\text{Case 1:} ~l \geq 0}$
Then, you can find an irrational $x_0$ arbitrarily close to $a$.
Then, $f(x_0) = -x_0 < \dfrac{-a}{2} \implies |l - f(x_0)| > \dfrac{a}{2}.$
$\underline{\text{Case 2:} ~l < 0}$
Then, you can find a rational $x_1$ arbitrarily close to $a$.
Then, $f(x_1) = x_1 > \dfrac{a}{2} \implies |l - f(x_1)| > \dfrac{a}{2}.$

Therefore, regardless of whether $l \geq 0,$ or $l < 0$, you can always find an element $x$ arbitrarily close to $a$ [in the two cases above, $x_0$ or $x_1$] such that $|f(x) - l| > \dfrac{a}{2}.$
The underlying idea is that both the rational numbers and irrational numbers are dense in $\Bbb{R}$, which signifies that given any $a \in \Bbb{R}$, you can find both a rational number and an irrational number arbitrarily close to $(a)$.
