<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] $$\sqrt{n^2+n} = n \sqrt{1 + \frac{1}{n} } = n (1 + \frac{1}{2n} + O(1/n^2)) =$$
$$ = n + \frac{1}{2} + O(1/n) $$ 
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