11. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1$, when $n \geqslant 2$, $a_{n} 、 S_{n} 、 S_{n}-\frac{1}{2}$ form a geometric sequence. Then $\lim _{n \rightarrow \infty} n^{2} a_{n}=$ $\qquad$ .
11. $-\frac{1}{2}$.

From the condition, when $n \geqslant 2$,
$$
S_{n}^{2}=a_{n}\left(S_{n}-\frac{1}{2}\right)=\left(S_{n}-S_{n-1}\right)\left(S_{n}-\frac{1}{2}\right) \text {. }
$$

Thus, $\frac{1}{S_{n}}-\frac{1}{S_{n-1}}=2$.
$$
\text { Then } \frac{1}{S_{n}}=\frac{1}{S_{1}}+2(n-1)=2 n-1 \text {. }
$$

Therefore, $S_{n}=\frac{1}{2 n-1}$.
So, $a_{n}=S_{n}-S_{n-1}$
$$
\begin{array}{l}
=\frac{1}{2 n-1}-\frac{1}{2 n-3} \\
=-\frac{2}{(2 n-1)(2 n-3)} .
\end{array}
$$

Hence, $\lim _{n \rightarrow \infty} n^{2} a_{n}=-\lim _{n \rightarrow \infty} \frac{2 n^{2}}{(2 n-1)(2 n-3)}$
$$
=-\lim _{n \rightarrow \infty} \frac{2}{\left(2-\frac{1}{n}\right)\left(2-\frac{3}{n}\right)}=-\frac{1}{2} .
$$