13. Given an arithmetic sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1$, and a positive geometric sequence $\left\{b_{n}\right\}$, the sums of the first $n$ terms of the arithmetic sequence $\left\{a_{n}\right\}$ and the geometric sequence $\left\{b_{n}\right\}$ are denoted as $A_{n}$ and $B_{n}$, respectively, satisfying $\lim _{n \rightarrow \infty}\left(\frac{a_{n}}{n}+b_{n}\right)=3$ and $\lim _{n \rightarrow \infty}\left(\frac{A_{n}}{n^{2}}+\frac{B_{n}}{n}\right)=2$.
（1）Prove: $\lim _{n \rightarrow \infty} b_{n}, \lim _{n \rightarrow \infty} \frac{B_{n}}{n}$ both exist;
（2）Find the expressions for $A_{n}, B_{n}$.
13. Let the common difference of the arithmetic sequence $\left\{a_{n}\right\}$ be $d$, and the common ratio of the geometric sequence $\left\{b_{n}\right\}$ be $q$,

then $a_{n}=1+(n-1) d, A_{n}=n+\frac{n(n-1)}{2} d$, we have $\frac{a_{n}}{n}=d-\frac{d-1}{n}, \frac{A_{n}}{n^{2}}=\frac{d}{2}-\frac{d-2}{2 n}$,
so $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=d, \lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty}\left(\frac{a_{n}}{n}+b_{n}-\frac{a_{n}}{n}\right)=3-d$;
Similarly, $\lim _{n \rightarrow \infty} \frac{A_{n}}{n^{2}}=\frac{d}{2}, \lim _{n \rightarrow \infty} \frac{B_{n}}{n}=2-\frac{d}{2}$, thus $\lim _{n \rightarrow \infty} \frac{B_{n}}{n}, \lim _{n \rightarrow \infty} b_{n}$ both exist and are bounded.
Let $b_{n}=b_{1} q^{n-1}\left(n \in \mathbf{N}^{*}\right)$ and $b_{1}>0, q>0$, since $\lim _{n \rightarrow \infty} b_{n}$ exists and is bounded, hence $0<q \leqslant 1$.
(1) When $0<q<1$, then $\lim _{n \rightarrow \infty} b_{n}=0, \lim _{n \rightarrow \infty} \frac{B_{n}}{n}=\frac{b_{1}}{1-q} \cdot \lim _{n \rightarrow \infty} \frac{1}{n}=0$, combining with $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=d, \lim _{n \rightarrow \infty} \frac{A_{n}}{n^{2}}=$ $\frac{d}{2}$ and the two given equations, we get $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=d=3$ and $\lim _{n \rightarrow \infty} \frac{A_{n}}{n^{2}}=\frac{d}{2}=2$, which is a contradiction.
(2) When $q=1$, $\lim _{n \rightarrow \infty} b_{n}=b_{1}, \lim _{n \rightarrow \infty} \frac{B_{n}}{n}=b_{1}$, combining with $\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=d, \lim _{n \rightarrow \infty} \frac{A_{n}}{n^{2}}=\frac{d}{2}$ and the two given equations, we get $d+b_{1}=3, \frac{d}{2}+b_{1}=2$, solving these gives $d=2, b_{1}=1$, hence $A_{n}=n^{2}, B_{n}=n\left(n \in \mathbf{N}^{*}\right)$.
In conclusion, $A_{n}=n^{2}, B_{n}=n\left(n \in \mathbf{N}^{*}\right)$ is the solution.