I agree with SlipEternal. On the other hand we can provide an argument that it's 0. (Clearly there will be other methods that say otherwise.)
Anyway, for very large n, and that $\displaystyle n^2 >> n$, the limit becomes $\displaystyle \lim_{n \to \infty} sin ^2 ( \pi \sqrt{n^2 + n} ) \to \lim_{n \to \infty} sin^2 (\pi n)$. Since n is an integer and $\displaystyle sin(\pi n) = 0$ for all integer values of n then:
$\displaystyle \lim_{n \to \infty} sin ^2 ( \pi \sqrt{n^2 + n} ) \to \lim_{n \to \infty} sin^2 (\pi n) = 0$.
-Dan
We can write
\begin{align} \lim_{n \to \infty} \sqrt{n^2+n} &= \lim_{n \to \infty} \sqrt{\tfrac14(4n^2 + 4n + 1-1)} \\ &= \lim_{n \to \infty}\tfrac12\sqrt{(2n+1)^2 - 1} \end{align}
So clearly, \begin{equation} \sqrt{n^2+n} \approx \tfrac12 (2n+1) \end{equation} and this is a very slight over-estimate for large $n$. (A better over-estimate is $\sqrt{n^2+n} \approx \tfrac12\left((2n+1)-\frac{1}{2(2n+1)}\right)$).
Thus $$\sin{\big(\pi\sqrt{n^2+n}\big)} \approx \sin {\big(\pi(n+\tfrac12)\big)}$$ and this means that $\sin{\big(\pi\sqrt{n^2+n}\big)}$ alternates between approximately $1$ and approximately $-1$. So there is no limit.
You'll find that my approximation of $\sqrt{n^2+n}$ is closer than topsquark's.
I find it unbelievable that this is even a multiple thread.
LOOK HERE.
It should be clear that there is no limit. Rather the function oscillates between 0 & 1.
More formally, we can demonstrate that $$\frac12\left( (2n+1) - \frac1{2n+1} \right) < \sqrt{n^2 + n} < \frac12(2n+1)$$ and then, perhaps via the identity $\sin{(A-B)} = \sin{(A)}\cos{(B)} - \cos{(A)}\sin{(B)}$, that $$\lim_{n \to \infty} \left|\sin{\left(\pi\sqrt{n^2+n}\right)}\right| = \left| \sin {\left(\tfrac12(2n+1)\pi\right)} \right| = 1$$