7. Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=2, a_{2}=6, \\
a_{n+1}=\frac{a_{n}^{2}-2 a_{n}}{a_{n-1}}(n=2,3, \cdots) .
\end{array}
$$

Then $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}+n}\right\}=$ $\qquad$
7.1 .

Notice that, $a_{1}=1 \times 2, a_{2}=2 \times 3$.
Assume $a_{n-1}=(n-1) n, a_{n}=n(n+1)$. Then
$$
\begin{array}{l}
a_{n+1}=\frac{n(n+1)(n(n+1)-2)}{(n-1) n} \\
=\frac{n(n+1)(n+2)(n-1)}{(n-1) n} \\
=(n+1)(n+2) .
\end{array}
$$

Thus, $a_{n}=n^{2}+n$ always holds.
Also, $\sqrt{n^{2}+2 n} \in(n, n+1)$, so
$\left\{\sqrt{a_{n}+n}\right\}=\left\{\sqrt{n^{2}+2 n}\right\}$
$=\sqrt{n^{2}+2 n}-n=\frac{2 n}{\sqrt{n^{2}+2 n}+n}$
$\Rightarrow \lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}+n}\right\}=\lim _{n \rightarrow \infty} \frac{2}{\sqrt{1+\frac{2}{n}}+1}=1$.