How to find such limits : $\lim_{m \to \infty} \lim_{n \to \infty} cos^{2m}(n! \pi x)$ How to find such limits : 
$$\lim_{m \to \infty} \lim_{n \to \infty}  cos^{2m}(n! \pi x)$$
Please suggest not getting any idea how to approach such problems, will be of great help thanks. 
 A: I saw a similar problem many years ago.  The idea is that any real number can be written in the form
$$ x = a_1 + \sum_{n=2}^\infty \frac{a_n}{n!} ,$$
where $a_1$ is an integer, and $a_n$ is an integer between $0$ and $n-1$.  Then
$$ \cos^2(n! \pi x) = \cos^2\left(\pi \left(\frac{a_{n+1}}{n+1} + O(\tfrac1{n})\right)\right) .$$
Hence the behavior of $\cos^2(n! \pi x)$ depends upon how $\frac{a_{n}}{n}$ behaves (modulo 1) as $n \to \infty$.  In this manner, by the appropriate choice of $x$, you can make
$$ f_x = \lim_{n\to\infty} \cos^2(n!\pi x) $$
be any number in $[0,1]$ you like, or you can even make the limit not exist.
And now you just have to worry about $\lim_{m\to\infty} f_x^m$.
It will converge to $1$ if and only if $\frac{a_{n}} n$ converges to $0$ modulo 1.  For example, if $x = e$, then the limit is $1$.
A: Here $x$ is a constant. It can either rational or irrational.


*

*$x$ is rational.
$x$ can be expressed as $\frac pq$ where $p$ and $q$ are natural numbers.
$n!$ is $1.2.3.4... n$ and as $n$ tends to infinity, $n!$ must contain $q$. Therefore $q$ cancels off  and all that is left is a positive integral multiple of $\pi$. 
$\cos k\pi$ is 1 or -1. So $\cos^2 k\pi$ is 1. Therefore the answer is 1.

*$x$ is irrational
Similar to the above case but the argument of the cosine function will be less            than $\pi$. So the range is -1 to 1. But since $\cos^2$, the range is 0 to 1.
Any number in 0 to 1 raised to infinity is 0. Therefore the answer is 0.
