Does the note matter?

Note first that

$\sqrt{n^{2} + n} = \sqrt{(n + \dfrac{1}{2})^{2} - \dfrac{1}{4}} = \dfrac{2n + 1}{2}\sqrt{1 - \dfrac{1}{(2n + 1)^{2}}} =$

$\left(n + \dfrac{1}{2}\right)\sqrt{1 - \dfrac{1}{(2n + 1)^{2}}}.$

In the limit as $n \rightarrow \infty$ this expression goes to $\left(n + \dfrac{1}{2}\right),$ and so

$\lim_{n \rightarrow \infty} \cos(\pi\sqrt{n^{2} + n}) = \cos(\pi * (\lim_{n \rightarrow \infty} \sqrt{n^{2} + n})) =$

$\lim_{n \rightarrow \infty} \cos(\pi*\left(n + \dfrac{1}{2}\right)) = \boxed{0}$

since for any integer $n$ we have $\cos(\pi\left(n + \dfrac{1}{2}\right)) = 0.$

Ok,so here is the shorter solution

First, take $n^2$ common from the root

$\displaystyle \lim_{n \to \infty } cos(\pi n \left( 1 + \dfrac{1}{n} \right)^{\frac{1}{2}})$

Now, I use this property the most

$(1+ x)^n = 1 + nx, x \ll 1$

Here

$\displaystyle \lim_{n \to \infty } cos( \pi n \left( 1 + \dfrac{1}{2n} \right) ) = \lim_{n \to \infty } sin( n\pi)$

Since n is an integer $\displaystyle \lim_{n \to \infty } sin( n \pi) = 0$ .

This proof does not rely on unexplained exchanges of limites. First examine the argument:

$\begin{aligned} \pi\sqrt{n^2+n} &= \pi(\sqrt{n^2+n}-n)+n\pi\underset{\text{3. Binomi}}{=}\pi\frac{n^2+n - n^2}{\sqrt{n^2+n}+n}+n\pi=\pi\frac{1}{\sqrt{1+n^{-2}}+1}+n\pi\qquad\left| \text{first part }\rightarrow\:\frac{\pi}{2} \right.\\\\ &=\pi\left( \frac{1}{\sqrt{1+n^{-2}}+1}-\frac{1}{2} \right)+\left(n+\frac{1}{2}\right)\pi=:c_n+\frac{2n+1}{2}\pi,\qquad |c_n|\rightarrow 0\quad\text{for }n\rightarrow\infty \end{aligned}$

Use that to gain an upper estimate of the given sequence:

$\begin{aligned} 0\leq |a_n|&=\left| \cos\left(\frac{2n+1}{2}\pi+c_n\right) \right|=\left| (-1)^n\cos\left(\frac{\pi}{2}+c_n\right) \right|=|-\sin(c_n)|=\sin|c_n|\rightarrow 0\\\\ \Rightarrow \quad a_n&\rightarrow\fbox{0} \end{aligned}$

Rem.: In the last step, we use that $\sin(x)$ is continuous in $x=0$ so that we can be sure the following limit exists:

$\begin{aligned}\lim_{x\rightarrow 0}\sin(x)=0\end{aligned}$

Believe it or not I just used the squeeze theorem on this problem, and i got it right...