14. Container $A$ contains $a$ liters of brine solution with a concentration of $p$, and container $B$ contains $b$ liters of brine solution with a concentration of $q$ $(p \neq q)$. Now, $m$ liters of solution are simultaneously taken from both $A$ and $B$ and poured into the other $(m<a, m<b)$, and this operation is repeated $n$ times. Find the concentrations $p_{n}, q_{n}$ of the brine solutions in $A$ and $B$, and determine $\lim _{n \rightarrow \infty} p_{n}$ and $\lim _{n \rightarrow \infty} q_{n}$.
14. According to the problem, we have
$$
\begin{array}{l}
p_{n+1}=\frac{p_{n}(a-m)+q_{n} \cdot m}{a}, \\
q_{n+1}=\frac{q_{n}(\underline{b}-m)+P_{n} \cdot m}{b}
\end{array}
$$

Then
$$
P_{n+1}-q_{n+1}=\frac{a b-a m-b m}{a b}\left(p_{n}-q_{n}\right) .
$$

For convenience, let $p_{0}=p, q_{0}=q$. The sequence $\left\{p_{n}-q_{n}\right\}$ is a geometric sequence with a common ratio of
$$
\frac{a b-a m-b m}{a b}
$$

Thus,
$$
\begin{array}{l}
p_{n}-q_{n}=\left(p_{0}-q_{0}\right) \cdot\left(\frac{a b-a m-b m}{a b}\right)^{n} \\
=(p-q)\left(\frac{a b-a m-b m}{a b}\right)^{n} \text {. } \\
\end{array}
$$

That is,
$$
q_{n}=p_{n}-(p-q)\left(\frac{a b-a m-b m}{a b}\right)^{n} .
$$

Substituting (2) into (1) and simplifying, we get
$$
p_{\pi+1}-p_{n}=-\frac{m(p-q)}{a} \cdot\left(\frac{a b-a m-b m}{a b}\right)^{n},
$$

Then $p_{n}=p+\left[\left(p_{1}-p\right)+\left(p_{2}-p_{1}\right)+\cdots+\left(p_{n}-p_{n-1}\right)\right]$
$$
\begin{array}{l}
=p-\frac{m(p-q)}{a} \cdot \frac{1-\left(\frac{a b-a m-b m}{a b}\right)^{n}}{1-\frac{a b-a m-b m}{a b}} \\
=\frac{a p+b q}{a+b}+\frac{p-q}{a+b} \cdot\left(\frac{a b-a m-b m}{a b}\right)^{\pi} .
\end{array}
$$

Similarly,
$$
q_{n}=\frac{a p+b q}{a+b}-\frac{p-q}{a+b} \cdot\left(\frac{a b-a m-b m}{a b}\right)^{n} .
$$

When $m=\frac{a b}{a+b}$, $\frac{a b-a m-b m}{a b}=0$, then for any $n$, we have $p_{n}=q_{n}=$ $\frac{a p+b q}{a+b}$

When $m \neq \frac{a b}{a+b}$, since $m<a, m<b$, it is easy to prove that $\left|\frac{a b-a m-b m}{a b}\right|<1, \lim _{n \rightarrow \infty} p_{n}=$ $\lim _{n \rightarrow \infty} q_{n}=\frac{a p+b a}{a+b}$.