kill(all)$

van=facebook.com/LearningCalculus$

f(x,m,n):=cos(n!*%pi*x)^(2*m); 

assume(m>0,n>0)$

L:limit(f(x,m,n),n,inf);

limit(L,m,inf);

10!;

diff(f(x,m,n),m);

The limit does not exist. 
Here is why:
cos(u) is a bounded function.
No matter what u is, the cos(u) is some value between -1 and 1 inclusive.
I will denote this idea of a closed interval with square braces,
as in the closed interval [-1,1].
If we raise any value in the closed interval to the power 2*m
we map values in that interval to the closed interval [0,1]
assuming m is some positive integer.
This is because any number raised to an even power is positive.
Since the value of x is independent of m and n given any m and n,
the function will have different values depending on the value of x.
If we require Limits to be single valued and we typically do,
then the limit of this equation does not exist.
If we allowed a more general notion of a bounded interval,
we could say that the value of this function must lie in the closed interval [0,1].
But that is all we can say.
Classic calculus would say the limit does not exist.
We are asked to calculate a value which does not exist.
But we can say for any x the value must lie in [0,1].
Note that infinity is not a number,
it a concept and there are different sizes of infinity,
so it is not a single valued concept.
We have a loss of closure since the type of thing a number is,
is not type of thing an infinity is
since a number has a single value and an infinity can have multiple values.
Another way of seeing this is that if we have cos(w u),
w is the frequency of the function. Since n goes to infinity,
the factorial of n goes to infinity,
and it does so faster than n itself.
So for infinite values of n, the function is changing infinity rapidly.
One might then say it BECOMES the interval [0,1]
since it is traversing all the contained values infinitely rapidly.