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? 

Assuming that you want to know more than the limit itself.
Using $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n}))$$ use Taylor series 
$$\sqrt{n^2 + n}=n+\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$
$$2\pi\sqrt{n^2 + n}=(2n+1)\pi -\frac{\pi}{4 n}+O\left(\frac{1}{n^2}\right)$$
$$\cos(2 \pi \sqrt{n^2 + n})\sim -\cos \left(\frac{\pi }{4 n}\right)=-1+\frac{\pi ^2}{32 n^2}+O\left(\frac{1}{n^4}\right)$$
$$\sin ^2 (\pi \sqrt{n^2 + n}) =1-\frac{\pi ^2}{64 n^2}+O\left(\frac{1}{n^4}\right)$$ then the limit and how it is approached.
Using $n=1000$, the "exact" value would be $$\sin ^2\left(10 \sqrt{10010} \pi \right)\approx 0.99999984594$$ while the above approximation would give
$$1-\frac{\pi ^2}{64000000}\approx 0.99999984578$$
