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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?

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    You seem to be mixing up two things: $\lim\limits_{n\to\infty}\cos\pi n!,\;n\in\mathbb R$ and $\lim\limits_{n\to\infty}\cos\pi n!,\;n\in\mathbb N$. The former is not defined because of discontinuity and stuff. The latter is obviously 1. – 2017-01-18

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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.

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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!