6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=6, a_{2}=20, a_{n}\left(a_{n}-8\right)=a_{n-1} a_{n+1}-12(n \geqslant 2) \text {. }
$$

Let $\{x\}=x-[x]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}=$ $\qquad$
6. $\frac{1}{2}$.

According to the recursive relation and initial value, it is easy to calculate $a_{3}=42$.
From $a_{n}\left(a_{n}-8\right)=a_{n-1} a_{n+1}-12$,
we get
$$
a_{n+1}\left(a_{n+1}-8\right)=a_{n} a_{n+2}-12 \text {. }
$$
Subtracting (1) from (2) gives
$$
a_{n+1}^{2}-a_{n}^{2}-8\left(a_{n+1}-a_{n}\right)=a_{n} a_{n+2}-a_{n-1} a_{n+1} \text {. }
$$

Rearranging gives
$$
\begin{array}{l}
a_{n}\left(a_{n+2}+a_{n}-8\right)=a_{n+1}\left(a_{n+1}+a_{n-1}-8\right) . \\
\text { Then } \frac{a_{n+2}+a_{n}-8}{a_{n+1}}=\frac{a_{n+1}+a_{n-1}-8}{a_{n}} \\
=\cdots=\frac{a_{3}+a_{1}-8}{a_{2}}=2,
\end{array}
$$

which means $a_{n+2}-2 a_{n+1}+a_{n}=8$.
Therefore, $\left(a_{n+2}-a_{n+1}\right)-\left(a_{n+1}-a_{n}\right)=8$.
Hence, $\left\{a_{n+1}-a_{n}\right\}$ is an arithmetic sequence with the first term $a_{2}-a_{1}=14$ and common difference 8, i.e.,
$$
\begin{array}{l}
a_{n+1}-a_{n}=14+8(n-1)=8 n+6 . \\
\text { Then } a_{n}=a_{1}+\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)=6+\sum_{i=1}^{n-1}(8 i+6) \\
=6+8 \cdot \frac{n(n-1)}{2}+6(n-1)=4 n^{2}+2 n .
\end{array}
$$

Since $(2 n)^{2}<4 n^{2}+2 n<(2 n+1)^{2}$, we have $\left[\sqrt{a_{n}}\right]=$ $2 n$. Thus,
$$
\begin{array}{l}
\left\{\sqrt{a_{n}}\right\}=\sqrt{4 n^{2}+2 n}-2 n \\
=\frac{2 n}{\sqrt{4 n^{2}+2 n}+2 n}=\frac{2}{\sqrt{4+\frac{2}{n}}+2} .
\end{array}
$$

Therefore, $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}=\frac{1}{2}$.