7. If $n \in \mathbf{N}^{*}$, then $\lim _{n \rightarrow \infty} \sin ^{2}\left(\pi \sqrt{n^{2}+n}\right)=$ $\qquad$ (Contributed by Jian Weifeng)
7. 1 .

By the induced formula, we know
$$
\sin ^{2}\left(\pi \sqrt{n^{2}+n}\right)=\sin ^{2}\left(\left(\sqrt{n^{2}+n}-n\right) \pi\right)=\sin ^{2}\left(\frac{n}{\sqrt{n^{2}+n}+n} \pi\right) .
$$

Therefore,
$$
\begin{aligned}
\lim _{n \rightarrow \infty} \sin ^{2}\left(\pi \sqrt{n^{2}+n}\right) & =\lim _{n \rightarrow \infty} \sin ^{2}\left(\frac{n}{\sqrt{n^{2}+n}+n} \pi\right) \\
& =\lim _{n \rightarrow \infty} \sin ^{2}\left(\frac{1}{\sqrt{1+\frac{1}{n}}+1} \pi\right) \\
& =\sin ^{2}\left(\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} \pi\right)
\end{aligned}
$$
$$
=\sin ^{2} \frac{\pi}{2}=1
$$