Find the limit $\lim \limits_{n\to \infty }\cos \left(\pi\sqrt{n^{2}-n} \right)$ I'd love your help with finding the following limit:
$$\lim_{n\to \infty }\cos (\pi\sqrt{n^{2}-n}).$$
I was asked to find this limit, but honestly I believe that it doesn't exist.
According to Heine Theorem of limit of functions, I can choose two sequences:
$x_{k}=2\pi k$ and $y_{k}=2\pi k+\pi$ and notice that when I apply the function on both of them, I'll get -1 and 1, respectively.
Am I right?
Thank you again.
 A: Since a nice formal argument has been supplied by Davide Giraudo, I will allow myself the luxury of informality.
Let $n$ be a large positive integer.
Complete the square. We have 
$$n^2-n=\left(n-\frac{1}{2}\right)^2 -\frac{1}{4}$$
Take the square root.  When $n$ is very large, the term $-1/4$ makes a vanishingly small contribution to the square root.
So our square root is nearly equal to $n-1/2$. And the cosine of $\pi n -\pi/2$ is $0$.
A: We have 
\begin{align*}
\cos (\pi\sqrt{n^2-n})&= (-1)^n\cos(\pi(\sqrt{n^2-n}-n))\\
&= (-1)^n \cos\pi\frac{-n}{\sqrt{n^2-n}+n}\\
&=(-1)^n\cos \pi\frac 1{\sqrt{1-\frac 1n }+1},
\end{align*}
hence $|\cos(\pi\sqrt{n^2-n})| = \left|\cos \left(\pi\frac 1{\sqrt{1-\frac 1n }+1}\right)\right|$. 
Since $\lim \limits_{n\to +\infty}\pi\frac 1{\sqrt{1-\frac 1n }+1} =\frac{\pi}2$, the $\cos$ is continuous and $\cos \frac{\pi}2 =0$ we conclude that the limit is $0$.
A: Considering the form $\cos(\pi n \sqrt{1-\frac{1}{n}})$ and using
Taylor's expansion for $\sqrt{1-\frac{1}{x}}$ $\to$ see here,
we get that when n is large $\cos (\pi\sqrt{n^{2}-n}) \approx \cos( \pi n -\frac{\pi}{2})$. Therefore, $L=0$.
Q.E.D.
