Limits of factorial function at infinity In a question to find limit of $\cos^2 (\pi n!) $ as $n$ tends to infinity. (Sorry for not using MathJax, typing on a phone)
It is assumed that factorial is only defined for positive integers and not using the real number extension gamma function.
I believe it should not be defined (since $n!$ is discontinuous) but answer given is $1$, based on the fact that it is always cosine of an integral multiple of $\pi$.
So is it valid to say that since in its domain, the function always has value $1$, its limit is $1$, not taking into account the neighborhood limit existence condition?
 A: According to tradition, when you write
$$\lim_{n\to \infty}\,f(n)$$
you are usually assumed to take the limit when $n$ goes to infinity taking integer values. In other words, you are calculating the limit of the sequence $\{f(n)\}$, i.e., the number $L$ such that for all $\varepsilon>0$, there is some integer $N_\varepsilon$ such that for all integers $n\geq N_\varepsilon$, $|f(n)-L|<\varepsilon$. Hence you don't care about the value $f(x)$ when $x$ is not a positive integer.
However, when one writes 
$$\lim_{x\to \infty}\,f(x)$$
or
$$\lim_{t\to \infty}\,f(t)\text{,}$$
you are usually assumed to take the limit when $x$ or $t$ goes to infinity taking real values. This means the real number $L$ such that for all $\varepsilon>0$, there is some real number $C_\varepsilon$ such that for all real numbers $x\geq C_\varepsilon$, $|f(x)-L|<\varepsilon$. Hence you have to take into account all values of $f(x)$ as $x$ takes arbitrary real values.
In conclusion, it is just a matter of implicit meaning, based on tradition, in the use of the variables to denote a variable that take integer values or a variable that takes real values.
A: In general, when you use n as a variable, which is a natural number, and you are looking for the value of n going to infinity, you're doing something different than looking at the limit of a function. Because of course, if you put the gamma function instead of the factorial, it's going to behave in a nasty way.
What you're looking for here, is the limit of a sequence, the one value to which you get arbitrarily close as n goes to infinity. In this case, when n is bigger than 1, n! is always going to be even, so the nth term of the sequence is 1 for all n bigger than 1, and the limit of a constant sequence is trivially the value of that sequence.
Edit: I now notice the ^2, so you don't even have to wait for the second term, it's constant from the beginning!
