$$\lim_{n\rightarrow \infty\\m\rightarrow \infty}cos^2 (n!πx) ^m$$
where n belongs to natural numbers and x belongs to real numbers. I have been told to find the graph of this function with variation of x
The only thing I have figured out is that the $cos^2 {x}$ has a range from [0 to 1] so the function will fluctuate between these values but I can't figure out how it will vary with x.
we have $\cos^{2m}(y)\in[0,1]$ because $2m$ is even. taking the limit we have $\cos^{2m}(y)\to0$ if $\cos(y)\ne\pm1$ and $\cos^{2m}(y)\to1$ if $\cos(y)=\pm1$.
Now, $\cos(y)=\pm1$ iff $y=N\pi,N\in\Bbb Z$ so $\lim_{m\to\infty}\cos^{2m}(y)=\begin{cases}1&y=N\pi,N\in\Bbb Z\\0&\mbox{otherwise}\end{cases}$.
Now $y=n!\pi x$, so we need $n!x$ to be an integer, $n!x\in\Bbb Z\iff x=\frac \beta{\alpha}$ where both are integers and $\alpha$ divides $n!$(and, ofc, $\alpha\ne0)$, we can notice that if $|\alpha|\le n$ it divides $n!$, taking the limit we have that for every integer $\alpha$.
Thus we left with $$f(x)=\begin{cases}1&x=\frac{\beta}{\alpha},\beta,\alpha_{\ne0}\in\Bbb Z\\0&\mbox{otherwise}\end{cases}$$
In other words
$$f(x)=\begin{cases}1&x\in\Bbb Q\\0&\mbox{otherwise}\end{cases}$$