For $x$ being irrational, why is $\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!\pi x)]^{2m}=0$? 
Let $x \in \mathbb{R} \setminus \mathbb{Q}$. What is the value of $$\lim_{m \to \infty} \lim_{n \to \infty} \left[ \cos(n!\pi x) \right]^{2m}, \qquad (m,n \in \mathbb{N})$$


The answer given is $0$. I don't understand why it is so. 
 A: Let $e=\sum_{j=0}^{\infty}j!^{-1},$ which is irrational, because $n!e\not \in \mathbb Z$ for any $n\in \mathbb N.$ 
For $n\in \mathbb N$ we have $$n!e=A_n+\sum_{j=n+1}^{\infty}n!/j!=A_n+B_n,$$ where $A_n\in \mathbb N$  and $B_n\in (0,1)$.By induction on $n,$ if $n$ is even then $A_n$ is odd, and if $n$ is odd then $A_n$ is even. 
So for infinitely many $n$ we have $\cos (\pi n!e)=\cos \pi B_n$ and for infinitely many $n$ we have $\cos (\pi n!e)=-\cos \pi B_n.$
Therefore, since $\lim_{n\to \infty }B_n=0,$ the sequence $(\cos n!\pi e)_{n\in \mathbb N}$  has a  $\lim \sup$ of $+1$ and a $\lim \inf$ of $-1.$ Therefore  $\lim_{n\to \infty}\cos(n!\pi e)$ does not exist.
A: [This answer has been rewritten thanks to the helpful criticism in Alex's comment.]
The statement is wrong. One has to take into consideration first the existence of the limit
$$
\lim_{n\to\infty}\cos(n!\pi x)^{2m},
$$
which is a highly non-trivial problem. 
When $x=\dfrac{1}{\pi}$,
$
\cos(n!\pi x)=\cos(n!). 
$ But it seems that the existence of limit (which could be $1$)
$$
\lim_{n\to\infty}\cos(n!)\qquad\text{(which could be $1$)},
$$
is an open problem (edited due to Fimpellizieri's comment) thanks to the answer of this question: Is there a limit of cos (n!)?. 
[Added: Interestingly, there is  related question in MO: On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational.]

If you were to ask the double limit related to the Dirichlet function on the other hand:
$$
\lim_{m\to\infty}\lim_{n\to\infty}\big[\cos(m!\pi x)\big]^{2n}.
$$ 
you could read the accepted answer to this question:
Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals.
Note carefully (again, thanks to Alex's comment) that the order of taking the double limit is different from the one in your question:
$$
\lim_{m\to\infty}\lim_{n\to\infty}\big[\cos(m!\pi x)\big]^{2n}.
$$
