<s>[INST] Evaluate $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n})$ I'm struggling to find the limit at infinity of : $$\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}), n\in\Bbb N$$
I know it is $1$ but I don't understand why this is wrong :
$\sin^2(\pi\sqrt{n^2 + n}) = \sin^2(\pi*n\sqrt{1 + \frac{1}{n}})$.
So $\lim \limits_{n\rightarrow\infty}\sin^2(\pi\sqrt{n^2 + n}) = \lim \limits_{n\rightarrow\infty}\sin^2(\pi*n\sqrt{1 + \frac{1}{n}}) =  \lim \limits_{n\rightarrow\infty} \sin^2(\pi*n) = 0$.
Thanks.
 [/INST] First of all, notice that the function $f (\theta) = \sin^2 \theta$ is periodic with a period $\pi$. Then, for any integer $n$ we should have $\sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \left (\pi \left(\sqrt {n^2 + n} - n\right)\right)$. Since $$\lim_{n \to \infty} \left(\sqrt {n^2 + n} - n\right) = \frac {1} {2},$$ it becomes obvious that $$\lim_{n \to \infty} \sin^2 (\pi \sqrt {n^2 + n}) = \sin^2 \frac {\pi} {2} = 1.$$
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