Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x) = \left\{ \begin{array}{ll} 2x & \quad x \text{ is rational} \\ -2x & \quad x \text{ is irrational } \end{array} \right.$
Show that $f$ is discontinuous at every point $x_0 \neq 0$.
My idea to show this is to consider when $x_0$ is rational and then consider when it is irrational. Then using the density of the rationals we have sequences which converge to $x_0$. Then use The Divergence Criterion for Functional Limits Theorem to finish the proof. Is this along the right track?
Let $x_0 \in \mathbb R$. Take a sequence $(r_n)$ of rational numbers and a sequence $(s_n)$ of irrational numbers with $r_n \to x_0$ and $s_n \to x_0$.
Then: $f(r_n) =2r_n \to 2x_0$ and $f(s_n) =-2s_n \to -2x_0$.
This shows that $f$ is not continuous in $x_0$, if $x_0 \ne 0.$