14. As shown in Figure 4, there is a sequence of curves $P_{0}, P_{1}, P_{2}, \cdots$, where it is known that the area of the equilateral triangle enclosed by $P_{0}$ is 1. $P_{\lambda+1}$ is obtained by performing the following operation on $P_{A}$: divide each side of $P_{k}$ into three equal parts, construct an equilateral triangle outward using the middle segment of each side as the base, and then remove the middle segment $(k=0,1,2, \cdots)$. Let $S_{n}$ be the area of the figure enclosed by the curve $P_{n}$.
(1) Find the general term formula for the sequence $\left\{S_{n}\right\}$;
(2) Find $\lim _{n \rightarrow \infty} S_{n}$.
14. (1) As shown in Figure 4, when operating on $P_{0}$, it is easy to see that each side of $P_{0}$ becomes 4 sides of $P_{1}$, so the number of sides of $P_{1}$ is $3 \times 4$; similarly, when operating on $P_{1}$, each side of $P_{1}$ becomes 4 sides of $P_{2}$, so the number of sides of $P_{2}$ is $3 \times 4^{2}$. Thus, it is easy to deduce that the number of sides of $P_{n}$ is $3 \times 4^{n}$.

Given that the area of $P_{0}$ is $S_{0}=1$. Comparing $P_{1}$ with $P_{0}$, it is easy to see that $P_{1}$ adds a small equilateral triangle to each side of $P_{0}$, with an area of $\frac{1}{3^{2}}$. Since $P_{0}$ has 3 sides, $S_{1}=S_{0}+3 \times \frac{1}{3^{2}}=1+\frac{1}{3}$. Comparing $P_{2}$ with $P_{1}$, we see that $P_{2}$ adds a small equilateral triangle to each side of $P_{1}$, with an area of $\frac{1}{3^{2}} \times \frac{1}{3^{2}}$. Since $P_{1}$ has 3 $\times 4$ sides, $S_{2}=S_{1}+3 \times 4 \times \frac{1}{3^{4}}=1+\frac{1}{3}+\frac{4}{3^{3}}$.
Similarly, $S_{3}=S_{2}+3 \times 4^{2} \times \frac{1}{3^{6}}=1+\frac{1}{3}+\frac{4}{3^{3}}+\frac{4^{2}}{3^{5}}$. Thus, we conjecture
$$
\begin{aligned}
S_{n} & =1+\frac{1}{3}+\frac{4}{3^{3}}+\frac{4^{2}}{3^{5}}+\cdots+\frac{4^{n-1}}{3^{2 n-1}} \\
& =1+\sum_{k=1}^{n} \frac{4^{k-1}}{3^{2 k-1}}=1+\frac{3}{4} \sum_{k=1}^{n}\left(\frac{4}{9}\right)^{k} \\
& =1+\frac{3}{4} \times \frac{\frac{4}{9}\left[1-\left(\frac{4}{9}\right)^{n}\right]}{1-\frac{4}{9}} \\
& =1+\frac{3}{5}\left[1-\left(\frac{4}{9}\right)^{n}\right]=\frac{8}{5}-\frac{3}{5} \times\left(\frac{4}{9}\right)^{n} .
\end{aligned}
$$

We now prove formula (1) using mathematical induction.
When $n=1$, it is known from above that formula (1) holds.
Assume that for $n=k$, $S_{k}=\frac{8}{5}-\frac{3}{5} \times\left(\frac{4}{9}\right)^{k}$.
When $n=k+1$, it is easy to see that after the $(k+1)$-th operation, comparing $P_{k+1}$ with $P_{k}$, $P_{k+1}$ adds a small equilateral triangle to each side of $P_{k}$, with an area of $\frac{1}{3^{2(k+1)}}$, and $P_{k}$ has $3 \times 4^{k}$ sides, so
$$
\begin{array}{l}
S_{k+1}=S_{k}+3 \times 4^{k} \cdot \frac{1}{3^{2(k+1)}} \\
=S_{k}+\frac{4^{k}}{3^{2 k+1}}=\frac{8}{5}-\frac{3}{5} \times\left(\frac{4}{9}\right)^{k+1} .
\end{array}
$$

In summary, by mathematical induction, formula (1) is proven.
(2) $\lim _{n \rightarrow+\infty} S_{n}=\lim _{n \rightarrow+\infty}\left[\frac{8}{5}-\frac{3}{5} \times\left(\frac{4}{9}\right)^{n}\right]=\frac{8}{5}$.