<s>[INST] How can I find the limit of the following sequence $\sin ^2 (\pi \sqrt{n^2 + n})$? How can I  find the limit of the following sequence:
 $$\sin ^2 (\pi \sqrt{n^2 + n})$$
I feel that I will use the identity $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n})), $$
But then what? how can I deal with the limit of $\cos (2 \pi \sqrt{n^2 + n})$? I know that $\cos (n\pi) = (-1)^n$, if $n$ is a positive integer but then what? 
 [/INST] Intuitively, $n^2+n$ is almost exactly halfway between $n^2$ and $(n+1)^2$, and so $\sqrt{n^2+n}$ is very close to a half-integer. This means $\sin(\pi\sqrt{n^2+n})$ should be very close to $\pm 1$, and in either case $\sin^2(\pi\sqrt{n^2+n})$ will be very nearly $1$.
More formally, we have
$$\sin^2(\pi\sqrt{n^2+n})=\sin^2[\pi(\sqrt{n^2+n}-n)]
$$
Now, multiplying by $1$ in the form $\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}$ gives
$$
\sin^2[\pi(\sqrt{n^2+n}-n)]=\sin^2\left(\pi\frac{n}{\sqrt{n^2+n}+n}\right)
$$
Since $\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}+n}=\frac{1}{2}$ and $\sin^2$ is continuous, this sequence has limit $\sin^2 \frac{\pi}{2}=1$.
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