I'm examining the following function:
$ f:x \mapsto \left\{ \begin{array}{ll} x & x \in \mathbb{Q} \\ 0 & x \in \mathbb{Q}^C\\ \end{array} \right. $
I've been told that the limit of this function only exists at $0$. I have proved that the limit does, in fact, exist at 0. I seem to be hitting a roadblock with proving that the limit doesn't exist at any other points. I'm frankly unsure how to approach the proof using epsilon-delta. (Edit: How do I prove that $|f(x)-L| > \epsilon$ for any arbitrary L?) Some guidance would be very appreciated.
Let $x_0\in\mathbb{R}\setminus\{0\}$. Take a sequence $(q_n)_{n\in\mathbb{N}}$ of rational numbers and a sequence $(x_n)_{n\in\mathbb{N}}$ of irrational numbers such that $\lim_{n\to\infty}q_n=\lim_{n\to\infty}x_n=x$. Then$$\lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}q_n=x,$$whereas $\lim_{n\to\infty}f(x_n)=0$. Therefore, $\lim_{x\to x_0}f(x)$ does not exist and so $f$ is not continuous at $x_0$.