14. As shown in the figure, there is a sequence of curves $P_{0}, P_{1}, P_{2}, \cdots$. It is known that the figure enclosed by $P_{0}$ is an equilateral triangle with an area of 1, and $P_{k+1}$ is obtained from $P_{k}$ by the following operation: 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 $\lim _{n \rightarrow \infty} S_{n}$.
(1) The number of sides of the $n$-th curve $P_{n}$ is $a_{n}=3 \cdot 4^{n-1}$, and the number of small triangles added compared to the $(n-1)$-th curve $P_{n-1}$ is $a_{n-1}$, with each small triangle having an area of $\frac{1}{9^{n-1}}$.
Thus, $S_{n}=S_{n-1}+3 \cdot 4^{n-2} \cdot \frac{1}{9^{n-1}}=S_{n-1}+\frac{3}{4} \cdot\left(\frac{4}{9}\right)^{n-1}(n \geqslant 2)$.
Therefore, $S_{n}=S_{1}+\left(S_{2}-S_{1}\right)+\left(S_{3}-S_{2}\right)+\cdots+\left(S_{n}-S_{n-1}\right)$ $=1+\frac{3}{4} \cdot \frac{4}{9}+\frac{3}{4} \cdot\left(\frac{4}{9}\right)^{2}+\cdots+\frac{3}{4} \cdot\left(\frac{4}{9}\right)^{n-1}=1+\frac{\frac{1}{3}\left[1-\left(\frac{4}{9}\right)^{n-1}\right]}{1-\frac{4}{9}}=\frac{8}{5}-\frac{3}{5} \cdot\left(\frac{4}{9}\right)^{n-1}$. For $n=1$, the formula also holds. Therefore, $S_{n}=\frac{8}{5}-\frac{3}{5} \cdot\left(\frac{4}{9}\right)^{n-1}\left(n \in \mathbf{N}^{*}\right)$.
(2) $\lim _{n \rightarrow \infty} S_{n}=\frac{8}{5}-\frac{3}{5} \lim _{n \rightarrow \infty}\left(\frac{4}{9}\right)^{n-1}=\frac{8}{5}$.