Question

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

• A. 2
• B. 1
• C. 0
• D. None of these

Answer 1

Answer: (A), (B)

We know that $|\cos \theta |\le 1$ for all $\theta$.
So, if $|\cos n!px|<1$,
$\underset{m\to \infty }{\lim }\underset{n\to \infty }{\lim }\left(1+{\cos }^{2m}n!\pi x\right)$
$=(1+0)=1$
and if $|\cos n!px|=1$,
$\underset{m\to \infty }{\lim }\underset{n\to \infty }{\lim }\left(1+{\cos }^{2m}n!\pi x\right)$
$=\underset{m\to \infty }{\lim }\underset{n\to \infty }{\lim }\left(1+1^{2m}\right)$
$=\underset{m\to \infty }{\lim }\underset{n\to \infty }{\lim }(1+1)=2$