$$
=\lim _{n \rightarrow+\infty} \frac{2 n}{\sqrt{4 n^{2}+2 n}+2 n}=\frac{1}{2} .
$$
10. From the given information,
$$
a_{n+1}^{2}-8 a_{n+1}=a_{n} a_{n+2}-12 \text {. }
$$

Combining with equation (1), we get
$$
\begin{array}{l}
a_{n}\left(a_{n}+a_{n+2}-8\right)=a_{n+1}\left(a_{n-1}+a_{n+1}-8\right) \\
\Rightarrow \frac{a_{n}+a_{n+2}-8}{a_{n+1}}=\frac{a_{n-1}+a_{n+1}-8}{a_{n}} .
\end{array}
$$

Thus, the sequence $\left\{\frac{a_{n}+a_{n+2}-8}{a_{n+1}}\right\}$ is a constant sequence.
$$
\begin{array}{l}
\text { Given } a_{1}=6, a_{2}=20, \\
a_{3}=\frac{a_{2}\left(a_{2}-8\right)+12}{a_{1}}=42,
\end{array}
$$

we have
$$
\begin{array}{l}
\frac{a_{1}+a_{3}-8}{a_{2}}=2 . \\
\text { Hence } \frac{a_{n}+a_{n+2}-8}{a_{n+1}}=2 \\
\Rightarrow a_{n+2}-2 a_{n+1}+a_{n}-8=0 \\
\Rightarrow a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)+8
\end{array}
$$

Also, $a_{2}-a_{1}=14$, then
$$
\begin{array}{l} 
a_{n+1}-a_{n}=6+8 n(n=1,2, \cdots) \\
\Rightarrow a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) \\
=6+\sum_{k=1}^{n-1}(6+8 k)=4 n^{2}+2 n .
\end{array}
$$

Since $(2 n)^{2}<a_{n}<(2 n+1)^{2}$, then
$$
\left[\sqrt{a_{n}}\right]=2 n \text {. }
$$

Therefore, $\lim _{n \rightarrow+\infty}\left\{\sqrt{a_{n}}\right\}=\lim _{n \rightarrow+\infty}\left(\sqrt{a_{n}}-\left[\sqrt{a_{n}}\right]\right)$
$$
=\lim _{n \rightarrow+\infty}\left(\sqrt{4 n^{2}+2 n}-2 n\right)
$$
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=6, a_{2}=20, \\
a_{n}\left(a_{n}-8\right)=a_{n-1} a_{n+1}-12 .
\end{array}
$$

Let $\{x\}=x-[x]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. Find the value of $\lim _{n \rightarrow+\infty}\left\{\sqrt{a_{n}}\right\}$.