Find the limit $\lim \limits_{n\to \infty }\cos \left(\pi\sqrt{n^{2}-n} \right)$

Question

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.

Answer 1

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 $0$.

Answer 2

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 $0$.

Answer 3

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.