Demonstration of an equivalence between a function a limit.

Question

Given f(x)={0 if x is irrational and 1 if x is rational} How do I prove the following equality is true? :

Answer 1

Hint: if $x$ is rational then $x=\frac{p}{q}$ when $p\in\mathbb{Z},q\in\mathbb{N}$. When $n\geq q$ we have that $n!$ is divisible by $q$. What is the value of $\cos(n!\pi x)$ then? Try to finish from here.

Now, if $x$ is irrational then for any $n\in\mathbb{N}$ we have $n!x\notin\mathbb{Z}$. Once again I'll leave you to think why is that true, try to prove by contradiction. And after you do that, what can you say about the value of $\cos(n!\pi x)$?