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? 

You can check $\sin^2(\pi\sqrt{n^2 + n})=\sin^2(\pi\sqrt{n^2 + n}-\pi n)$.
So $$\sin^2(\pi\sqrt{n^2 + n})=\sin^2 \pi\frac{n}{\sqrt{n^2 + n}+n}\to \sin^2\frac{\pi}{2}=1.$$
