Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$

Note that $$\sqrt{n^2+n+1}-n=\frac{n+1}{\sqrt{n^2+n+1}+n}\to 1/2$$ as $n\to\infty$.

For even $n$, $\sin(\sqrt{n^2+n+1}\pi)=\sin(\sqrt{n^2+n+1}\pi-n\pi)\to \sin(\pi/2)=1$ as $n\to\infty, n$ even.

For odd $n$, $\sin(\sqrt{n^2+n+1}\pi)=-\sin(\sqrt{n^2+n+1}\pi-n\pi)\to -\sin(\pi/2)=-1$ as $n\to\infty, n$ odd.

Therefore, $$|\sin(\sqrt{n^2+n+1}\pi)|\to 1$$ as $n\to\infty$.

Edit: This assumes that $n$ is restricted to the integers.

Firstly, since $ \mid \sin \theta \mid $ is periodic with period $\pi$, we want to look at the value of $ \pi \sqrt{n^2 + n +1} \pmod{\pi}$.

Secondly, convince yourself that $\lim_{n\rightarrow \infty} \pi \sqrt{n^2+n+1} \pmod{\pi} = \frac {1}{2} \pi $.

Thirdly, by the continuity of $\sin \theta$, $ \mid \lim_{n \rightarrow \infty} \sin (\pi \sqrt{n^2+n+1} ) \mid = \mid \sin \lim_{n\rightarrow \infty} \pi \sqrt{n^2+n+1}\mid = \sin \frac {1}{2} \pi$.

If you know a little bit about Taylor expansions and big O notation, you can do that easily. We will use $$ \sqrt{1+u}=1+\frac{u}{2}+O(u^2)\qquad \mbox{when } u\longrightarrow 0. $$ Now $$ \pi\sqrt{n^2+n+1}=\pi n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=\pi n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)=\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right) $$ So $$ \sin (\pi\sqrt{n^2+n+1})=\sin\left(\pi n+\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)=(-1)^n\sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right). $$ It follows that $$ |\sin (\pi\sqrt{n^2+n+1})|=\left| \sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)\right|\longrightarrow \left| \sin\left(\frac{\pi}{2}\right)\right|=1. $$