TITLE: $\lim_{n\to \infty }\cos (\pi\sqrt{n^{2}-n})$ - second battle
QUESTION [5 upvotes]: Followed my  question,
I still don't understand why the answers that were given are right.
By simply intuition and using the continuity of cosine, we get that we need to compute
$\lim_{n\to \infty}\cos (\pi n\sqrt{1-\frac{1}{n}})$ and because of that, as I see that, there are two clear limits: 1 and -1, and therefore the limit does not exist
Can someone please explain me why am I wrong? two answers that claimed that the limit is 0 got 25 votes together, so I must be mistaken, But still the answers are not satisfying me.
By the way, Wolfarmalpha claims it does not have a limit.
Thank you.

REPLY [3 votes]: The behavior of the sequence 
$$
\{\cos(\pi \sqrt{n^2-n})\}_{n=1}^\infty
$$ 
and that of the function 
$$
\cos(\pi \sqrt{x^2-x})
$$ 
are very different, which has been pointed out by other answers and comments. 
I would like to just add three pictures created with Mathematica to illustrate this point here: