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Q. $\displaystyle\lim _{m \rightarrow \infty} \displaystyle\lim _{n \rightarrow \infty}\left(1+\cos ^{2 m} n ! \pi x\right)$ is equal to

Limits and Derivatives

Solution:

We know that $|\cos \theta| \leq 1$ for all $\theta$.
So, if $|\cos n ! p x|<1$,
$\displaystyle\lim _{m \rightarrow \infty} \displaystyle\lim _{n \rightarrow \infty}\left(1+\cos ^{2 m} n ! \pi x\right)$
$=(1+0)=1$
and if $|\cos n ! p x|=1$,
$\displaystyle\lim _{m \rightarrow \infty} \displaystyle \lim _{n \rightarrow \infty}\left(1+\cos ^{2 m} n ! \pi x\right)$
$=\displaystyle\lim _{m \rightarrow \infty} \displaystyle\lim _{n \rightarrow \infty}\left(1+1^{2 m}\right)$
$=\displaystyle\lim _{m \rightarrow \infty} \displaystyle\lim _{n \rightarrow \infty}(1+1)=2$