<s>[INST] Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$ Calculate 
$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$
 [/INST] 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
$$
|\sin (\pi\sqrt{n^2+n+1})|=\left| \sin\left(\frac{\pi}{2}+O\left(\frac{1}{n} \right) \right)\right|\longrightarrow \left| \sin\left(\frac{\pi}{2}\right)\right|=1.
$$
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