Limit of Function series 
Given that $$g(x):=\lim _{ n\rightarrow \infty  }{ { (\cos(x)) }^{ 2n } } $$
  Find 
     $$h(x)=\lim _{ m\rightarrow \infty  }{ { g(x) }_{ m } } $$
  Where
  ${ g(x) }_{ m }:=g(2\pi m!x)$.


I already computed $g(x)=0$
but I can't compute $h(x)$ the same way I did with $g(x)$.
Could someone give me a hint on how to start computing $h(x)$?

Remember that:
$$\lim_{n\to \infty} x^{2n}=
 \begin{cases} 
      \infty & |x| > 1 \\
      1 & |x| = 1 \\
      0 & |x| < 1 
   \end{cases}
$$
In your case, $\cos x\le1$ . Note that $\cos(2\pi m!x)$ is a rather convoluted notation for the simpler $\cos(2\pi zx)$ where $z$ is an integer.
Now what is the value of $\cos(2\pi zx)$ always equal to? Can you now solve the limit?

Remark:

But what if $zx=1/4$, wouldn't $\cos(2\pi zx)=\cos (\pi/2)=0?$

That would not actually happen. Since $z=m!$, that implies that even if $x=p/q$ (in reduced form, $p,q\in Z$), that $q$ will get cancelled with $m!$, as $m!=1\cdot 2\cdot3...\cdot (q-1)\cdot q\cdot (q+1)....m$.
Because $m\to\infty$, we are covering all integers that could possibly ever exist (at least theoretically), hence we are sure to have $q$ get cancelled at some point, and the $2\pi$ would remain safe.
