Evaluating the double limit $\lim_{m \to \infty} \lim_{n \to \infty} \cos^{2m}(n! \pi x)$ I have to find out the following limit $$\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!πx)^{2m}]$$ for $x$ rational and irrational. for $x$ rational $x$ can be written as $\frac{p}{q}$ and as $n!$ will have $q$ as its factor the limit should be equal to 1.
the second part of irrational is giving me problems. I first thought that limit should be zero as absolute value of cosine term is less than 1 and power it to infinity you should get $0$. But then I realised that it was wrong. I brought the limit down to this form. $$e^{-\sin^2(n!πx)m}$$ after this I find the question quite ambiguous as they have just said $x$ is irrational. If I take $x$ as $\frac{1}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e}$ but if I take  $x$ as $\frac{2}{n!π\sqrt{m}}$ I get the limit as $\frac{1}{e^4}$. please help me and tell me where have I gone wrong?

For $m>0$ the limit of $cos^{2m}(n!\pi.x)$ as $n$ goes to infinity does not exist for some $x$. For example, for natural number k, let $f(k)=1/2$ if $k$ is an integer power of $2$, otherwise let $f(k)=0$. Let $x=f(0)/0!+f(1)/1!+f(2)/2!+...$
