Limit of a sequence in dependency to x [closed]

Calculate the limit of $$\lim\limits_{n\to\infty}\lim\limits_{k\to\infty}(cos(n!x))^{2k} $$ in dependency to x $$\\$$ Since $$\lim\limits_{k\to\infty}(cos(n!x))^{2k} = 0\\ $$ I don't know how can I continue.

We only have $$\lim_{k\to\infty} (\cos(n!x))^{2k}=0$$ if $$|\cos(n!x)|<1.$$

This will only not be the case when $\cos(n!x)=1$ or $\cos(n!x)=-1$. We solve this to get $$n!x=2\pi j \;\;\text{and}\;\; n!x=\pi+2\pi j$$ for $j\in \mathbb Z$. I.e. $$x=\frac{2\pi}{n!}j \;\;\text{and}\;\; x=\frac{\pi+2\pi}{n!} j.$$ And since $n \to \infty$ we must have that $x=0$ is the only exception.

Therefore, $$ \lim_{n\to\infty}\lim_{k\to\infty} (\cos(n!x))^{2k} = \begin{cases} 1, & \text{if $x=0$} \\[2ex] 0, & \text{otherwise} \end{cases}$$