Show that the limit of the function does not exist at $x \neq 0 $ 
Consider the function $ f(x) = \left\{    \begin{array}{l l}
     x & \quad \text{if $x$ is irrational}\\
     -x & \quad \text{if $x$ is rational}   \end{array} \right.$
Show that $\lim_{x \to a} f(x)$ does not exist    by definition of
  $\epsilon  -\delta$ , where $ a \neq 0$ is a real number.

Let if possible $\lim_{x \to a} f(x) = l$, so for each $\epsilon > 0 $, there exist $\delta > 0 $ such that $| f(x) - l| < \epsilon$, whenever $0 < | x- a | < \delta $.
Please tell me how to solve further. 
 A: Let take a sequence $(a_n)_{n\in\mathbb N}\subset \mathbb Q$ and a sequence $(b_n)_{n\in\mathbb N}\subset \mathbb R\backslash \mathbb Q$ who both converge to $\ell\neq 0$.
You have that $\lim_{n\to\infty}f(a_n)=-\ell$ and $\lim_{n\to\infty }f(b_n)=\ell$.
Q.E.D
A: Let us suppose $a>0$. 
$l>0$ cannot be the limit, because you can take $0<\varepsilon<l$, then you should have $0<l-\varepsilon<f(x)<l+\varepsilon$ and for each $\delta<a$ you choose, the image $f(x)$ of a rational point is negative and does not fall in $(l-\varepsilon,l+\varepsilon)$.
$l<0$ cannot be the limit, because you can take $0<\varepsilon<-l$, then you should have $l-\varepsilon<f(x)<l+\varepsilon<0$ and for each $\delta<a$ you choose, the image $f(x)$ of an irrational point is positive and does not fall in $(l-\varepsilon,l+\varepsilon)$.
$l=0$ cannot be the limit, because you can take $0<\varepsilon<a$, then you should have $-\varepsilon<f(x)<+\varepsilon$ and for each $\delta<a-\varepsilon$ you choose, the image $f(x)$ of an irrational point is $f(x)=x>a-\delta>\varepsilon$ and does not fall in $(-\varepsilon,+\varepsilon)$, and the image $f(x)$ of a rational point is $f(x)=-x<-a+\delta<-\varepsilon$ and does not fall in $(-\varepsilon,+\varepsilon)$.
Similar reasonings are valid for $a<0$.
