How to solve double limit $\lim_{m\to\infty}[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}]$ Please provide some hints as to how to solve questions with double limits such as this:
$$\lim_{m\to\infty}\left[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}\right]$$
One of the things I did was convert the original function to:
$$e^{n\ln(\cos(m!\cdot\pi\cdot x)^2)}$$
and then change cosine into sine and take $t=1/m$, and try to use 
$$\lim \frac{\sin(m!\cdot\pi\cdot x)}{m!\cdot\pi\cdot x}$$
but that only messed it up even further. I obviously can't use L'Hopital as not both the numerator and denominator go to zero. 
Another thing was to try to use the power series expansion, but that seemed even more complicated as there is still the power of 2n to deal with. 
Please help! Thanks.
 A: Hint: Look separately at $x$ irrational, $x$ rational.
If $x$ is irrational, then $\cos(m!\pi x)$ has absolute value less than $1$.
If $x$ is rational, say $\frac{a}{b}$ where $a$ and $b$ are integers,  what is the value of  $\cos(m!\pi x)$ for large enough $m$? 
Detail: Suppose that $x$ is irrational. Fix $m$. Then $m!\pi x$ is not an integer multiple of $\pi$. It follows that $|\cos(m!\pi x)|\lt 1$. Let $c_m=\cos(m\pi x)$. Since $c_m$ has absolute value less than $1$, we have $\lim_{n\to\infty} c_m^{2n}=0$. Thus our double limit is $\lim_{m\to\infty} 0$, which is $0$.
Now suppose that $x$ is rational. Then we can assume that $x=\frac{a}{b}$, where $a$ and $b$ are integers. Without loss of generality we may suppose that $b$ is positive. 
Let $m\ge b$. Then $m!x=(b-1)!a$. Thus $m!\pi x$ is an integer multiple of $\pi$. It follows that $\cos(m!\pi x)=\pm 1$, and therefore $(\cos(m!\pi x))^{2n}=1$. Thus for any $m\ge b$, we have
$$\lim_{n\to\infty} (\cos(m!\pi x))^{2n}=1.$$
We conclude that 
$$\lim_{m\to\infty}\left[ \lim_{n\to\infty} (\cos(m!\pi x))^{2n}             \right]=1.$$
