If$f\left(x\right)$ is a real number in [0, 1], then the value of the function $f\left(x\right)=\underset{m\to \mathrm{\infty }}{lim} \underset{n\to \mathrm{\infty }}{lim} \left(1+{\cos }^{2m}(n!\pi x)\right)$ is
given by

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

$\text{ We know that, }g\left(x\right)=g(x+T)$

$g\left(bx\right)=\frac{T}{\left|b\right|}$

$\text{ let, }g\left(x\right)=\cos (n!\pi x)$

$\text{ Where, pepiod }=\frac{2\pi }{n!\pi }$

$\because g\left(x\right)\in [-1,1]$

 $\text{ Hence, the value of }f\left(x\right)\text{ is }\{1,2\}\text{. }$