(11 Given $(1+\sqrt{3})^{n}=a_{n}+b_{n} \sqrt{3}$, where $a_{n}, b_{n}$ are integers, then $\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}}=$
$11 \sqrt{3}$ Hint: From the condition $(1+\sqrt{3})^{n}=a_{n}+b_{n} \sqrt{3}$, we know
$$
(1-\sqrt{3})^{n}=a_{n}-b_{n} \sqrt{3},
$$

thus
$$
\begin{array}{l}
a_{n}=\frac{1}{2}\left[(1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}\right], \\
b_{n}=\frac{1}{2 \sqrt{3}}\left[(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}\right],
\end{array}
$$

therefore
$$
\begin{aligned}
\lim _{n \rightarrow+\infty} \frac{a_{n}}{b_{n}} & =\lim _{n \rightarrow+\infty} \sqrt{3} \times \frac{(1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}}{(1+\sqrt{3})^{n}-(1-\sqrt{3})^{n}} \\
& =\lim _{n \rightarrow+\infty} \sqrt{3} \times \frac{1+\left(\frac{1-\sqrt{3}}{1+\sqrt{3}}\right)^{n}}{1-\left(\frac{1-\sqrt{3}}{1+\sqrt{3}}\right)^{n}} \\
& =\sqrt{3} .
\end{aligned}
$$