TITLE: How to prove that $\lim(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k}))=\chi_\mathbb{Q}$
QUESTION [11 upvotes]: Possible Duplicate:
How is this called? Rationals and irrationals 

Please help me prove, that
$$\underset{n\rightarrow\infty}{\lim}\left(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k})\right)=\begin{cases}
1 & \iff x\in\mathbb{Q}\\
0 & \iff x\notin\mathbb{Q}
\end{cases}$$
Seems very complicated, but it's on calc I. I've tried use series expansions of cos, but it don't lead to answer. Thanks in advance!
Edit
Please don't use too much advanced techniques.

REPLY [9 votes]: Baby Rudin has this as an example.

REPLY [2 votes]: Hint: Show that if $x \in \mathbb{Q}$, then there exists some $N$ such that for $n > N$, $n! \pi x$ is an integer multiple of $2\pi$. Conclude that it tends to 1.
Show that if $x \not \in \mathbb{Q}$, then $\lim_{k\rightarrow \infty} [\cos (n! \pi x)]^{2k} = 0$.
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