Given a function $f: \mathbb{R} \mapsto \mathbb{R}$ with $f(x) = 4x$ for rational number $x$ and $f(x) = x + 6$ for irrational number $x$. If $E = \{x \in \mathbb{R}| f~\text{continues on}~x\}$, then determine all the limit points of $E$.
This question came from real analysis olympiad in my region. How should I solve this? I don't know the first approachment, but we have tried some ways by using a definition of limit points, but seems confusing.
Assuming the condition on $E$ is that $f$ is continuous at $x$, the only point in $E$ is $x=2$. We find that by pretending $x$ is both rational and irrational so $4x=6+x$. At any $x \neq 2$ the values of $f(x)$ nearby on rational and irrational points will differ by enough that $f$ is not continuous. We can take $\epsilon = \frac 12(4x-(6+x)$ and no $\delta$ can be found. For $x=2$ we can do an $\epsilon-\delta$ proof of continuity with $\delta = \epsilon/4$. A set of only one point has no limit points, so the set of limit points of $E$ is $\emptyset$