Calculate $ \lim_{n\to+\infty}\cos{\big{(}\pi \sqrt{n^2+n}\big{)}}$ I need to prove that 
 $$ \lim_{n\to+\infty}\cos{\big{(}\pi \sqrt{n^2+n}\big{)}}=1$$
But this is something that seems highly unlikely to have a limit, but I am probably wrong. Wolfram Alpha says "$0$ to $1$". What does that even mean? How to prove this?
If you could provide only a hint (if this limit exists) so I can finish the task myself.
 A: The limit is not $1$.
Hint: $n^2+n=\left(n+\frac{1}{2}\right)^2-\frac{1}{4}$.
Show that $$\pi\sqrt{n^2+n}-\pi\left(n+\frac12\right)\to 0.$$
Note that this is an example of an ambiguity in the $\lim$ notation.
If you had instead asked for $\lim_{x\to\infty} \cos\left(\pi\sqrt{x^2+x}\right)$, you'd have no limit. The continuous limit, rather than the sequence limit, is what Wolfram Alpha is computing.
A: $$
\begin{align}
&\cos\left(\pi\sqrt{n^2+n}\right)\\
&=\cos\left(\pi\left(n+\sqrt{n^2+n}-n\right)\right)\\
&=\cos\left(\pi\left(n+\frac{n}{\sqrt{n^2+n}+n}\right)\right)\\
&=\cos\left(\pi n\right)\cos\left(\pi\frac{n}{\sqrt{n^2+n}+n}\right)-\sin\left(\pi n\right)\sin\left(\pi\frac{n}{\sqrt{n^2+n}+n}\right)\\
&=(-1)^n\cos\left(\frac\pi{\sqrt{1+1/n}+1}\right)
\end{align}
$$
Since
$$
\lim_{n\to\infty}\cos\left(\frac\pi{\sqrt{1+1/n}+1}\right)=\cos(\pi/2)=0
$$
we have
$$
\lim_{n\to\infty}\cos\left(\pi\sqrt{n^2+n}\right)=0
$$
A: Since $$0\leq \left(n+\frac{1}{2}\right)-\sqrt{n^2+n} = \frac{\frac{1}{4}}{\left(n+\frac{1}{2}\right)+\sqrt{n^2+n}}\leq\frac{1}{8n}$$
we get that, as long as $n\to +\infty$, $\pi\sqrt{n^2+n}$ gets closer and closer to $\pi\left(n+\frac{1}{2}\right)$. 
Since $\cos\left(\pi\left(n+\frac{1}{2}\right)\right)=-\sin(\pi n)$ and the cosine is a continuous function, the limit is just $\color{red}{0}$.
A: When $n \to \infty$
$$\sqrt{n^2+n} \approx n + \frac{1}{2}$$
$$\cos\left(\pi\left(n+\frac{1}2\right)\right)=0.$$
