I know that Thomae's function is continuous at all irrational numbers, so there are functional limits for those point. However, I don't know about rational numbers...
Thank you!
I know that Thomae's function is continuous at all irrational numbers, so there are functional limits for those point. However, I don't know about rational numbers...
Thank you!
If $a$ is a rational number, and $f$ is Thomae's function, then $\lim\limits_{x \to a} f(x) = 0$, but $f(a) \ne 0$. Thus Thomae's function has a limit at every real number, but is continuous exactly at the irrational numbers.
The fact that $\lim\limits_{x \to a} f(x) = 0$ for all $a$ is mentioned here; the proof is essentially identical to the irrational case, which is proven here.