Question

$\underset{m\to \infty }{lim}\underset{n\to \infty }{lim}\{1+{\cos }^{2m}(n!\pi x\left)\right\}$.

Answer 1

Case 1

If $x\in Q$

We know $-1\le \cos x\le 1\Rightarrow 0\le {\cos }^{2m}x\le 1$

We can write $x=\frac{p}{q}$, where $p,q\in N$

$\Rightarrow {lim}_{n\to \infty }n!\pi \frac{p}{q}=k\pi$, where $k$ is a integer.

$\therefore {\cos }^{2m}k\pi =1$

We have $\underset{m\to \infty }{lim}\underset{n\to \infty }{lim}\{1+co{s}^{2m}(n!\pi x\left)\right\}=1+1=2$

Case 2

If $x$ is irrational

${lim}_{m\to \infty }{\cos }^{2m}\left(k\right)=0$, where $k$ is non integer value.

We have $\underset{m\to \infty }{lim}\underset{n\to \infty }{lim}\{1+co{s}^{2m}(n!\pi x\left)\right\}=1+0=1$