11. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
\begin{array}{l}
a_{1}=3, a_{2}=7, \\
a_{n-1}\left(a_{n-1}-2\right)=a_{n-2} a_{n}-4(n=3,4, \cdots) .
\end{array}
$$

Let $[x]$ denote the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$. Find the value of $\lim _{n \rightarrow \infty}\left\{\sqrt{a_{n}}\right\}$.
11. From $a_{n-1}^{2}-2 a_{n-1}=a_{n-2} a_{n}-4$,
$$
a_{n-1} a_{n+1}-4=a_{n}^{2}-2 a_{n} \text {, }
$$

we get
$$
\begin{array}{l}
a_{n-1}^{2}-2 a_{n-1}+a_{n-1} a_{n+1} \\
=a_{n}^{2}-2 a_{n}+a_{n} a_{n-2} \\
\Rightarrow a_{n-1}\left(a_{n-1}+a_{n+1}-2\right) \\
\quad=a_{n}\left(a_{n-2}+a_{n}-2\right) .
\end{array}
$$

Noting that, $a_{1} \neq 0, a_{2} \neq 0, n \geqslant 3$, we have
$$
\begin{array}{l}
a_{n} a_{n-2}=a_{n-1}^{2}-2 a_{n-1}+4 \\
=\left(a_{n-1}-1\right)^{2}+3 \neq 0 \\
\Rightarrow a_{n} \neq 0 .
\end{array}
$$

From equation (1)
$$
\begin{array}{l}
\Rightarrow \frac{a_{n-1}+a_{n+1}-2}{a_{n}}=\frac{a_{n-2}+a_{n}-2}{a_{n-1}} . \\
\text { Then }\left\{\frac{a_{n-2}+a_{n}-2}{a_{n-1}}\right\}(n \geqslant 3) \text { is a constant sequence. }
\end{array}
$$
$$
\begin{array}{l}
\text { Given } a_{1}=3, a_{2}=7 \\
\Rightarrow a_{3}=\frac{a_{2}^{2}-2 a_{2}+4}{a_{1}}=13 \\
\Rightarrow \frac{a_{1}+a_{3}-2}{a_{2}}=2 \\
\Rightarrow \frac{a_{n-2}+a_{n}-2}{a_{n-1}}=2 \\
\Rightarrow a_{n}-a_{n-1}=a_{n-1}-a_{n-2}+2 \text {. } \\
\text { Also, } a_{2}-a_{1}=4 \text {, then } \\
a_{n}-a_{n-1}=2 n(n=2,3, \cdots) .
\end{array}
$$

Thus $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+$
$$
\begin{aligned}
& \left(a_{n}-a_{n-1}\right) \\
= & a_{1}+4+6+\cdots+2 n \\
= & n^{2}+n+1 .
\end{aligned}
$$

From $n<\sqrt{a_{n}}<n+1$
$$
\Rightarrow\left[\sqrt{a_{n}}\right]=n,\left\{\sqrt{a_{n}}\right\}=\sqrt{n^{2}+n+1}-n \text {. }
$$
$$
\begin{array}{l}
\text { Therefore, the original expression }=\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}+n+1}-n\right) \\
=\lim _{n \rightarrow \infty} \frac{n+1}{\sqrt{n^{2}+n+1}+n} \\
=\lim _{n \rightarrow \infty} \frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^{2}}}+1}=\frac{1}{2} .
\end{array}
$$