Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$ Calculate 
$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$
 A: 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$.
A: 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
$$
|\sin (\pi\sqrt{n^2+n+1})|=\left| \sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)\right|\longrightarrow \left| \sin\left(\frac{\pi}{2}\right)\right|=1.
$$
A: The function $|\sin x|$ is periodic with period $\pi$. Hence
$$\lim \limits_{n \to \infty}|\sin \pi \sqrt{n^2+n+1}|= \lim \limits_{n \to \infty}|\sin \pi (\sqrt{n^2+n+1}-n)|$$
But
$$\lim \limits_{n \to \infty} (\sqrt{n^2+n+1}-n)=\lim \limits_{n \to \infty} \dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}=\dfrac{1}{2}$$
It follows that the limit we are computing is equal to $\left |\sin \dfrac{\pi}{2}\right |$, which is 1.
A: We have
$$\sqrt{n^2+n+1}=n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=n\left(1+\frac{1}{2n}+o(\frac{1}{n})\right)=n+\frac{1}{2}+o(1)$$
hence
$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|=\lim \limits_{n \to \infty}|\sin(n\pi+\frac{\pi}{2}+o(1))|=\lim \limits_{n \to \infty}|\cos(o(1))|=1.$$
A: 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 $n\to\infty$.
