Let $$ f:[0,1]\to [0,1]\\ f(x)= \begin{cases} x, & x\notin\Bbb Q\\ 0, & x\in \Bbb Q \end{cases} $$
Prove the $\lim\limits_{x\to x_0}f(x)$ does not exist for $x_0 \in (0,1]$.
Attempt:
I don't know how to do this using the $\varepsilon-\delta$ definition, I tried supposing $\lim f(x)= \ell$ and going for a contradiction, but I couldn't reach any, I separated in the cases $\ell=0, \ell\neq 0$, but I couldn't get far with the $\ell\neq 0$ .
Could someone give me some hints/instructions on how to prove this? I'd like to solve it myself (at least as much as I can), rather than getting a complete solution.
HINT: If $L \neq 0$ try to take rational number between $x_0$ and $x_0+\delta$ and $\epsilon = \frac{|L|}{2}$
If $L=0 $ try to take irrational number between $x_0$ and $x_0+\delta$ and $\epsilon = x_0$