Example 29 (2002 National High School Competition Question) As shown in Figure 5-1, 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 by performing the following operation on $P_{k}$: divide each side of $P_{k}$ into three equal parts, use the middle segment of each side as the side to construct an equilateral triangle outward, 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 of the sequence $\left\{S_{n}\right\}$.
(2) Find $\lim _{n \rightarrow+\infty} S_{n}$.
Analysis Given $S_{0}=1$, it is not difficult to find $S_{n}$ and $\lim _{n \rightarrow+\infty} S_{n}$ by establishing the recursive relationship between $S_{n}$ and $S_{n-1}$.
Solution (1) Since $P_{0}$ has 3 sides, and each side of $P_{0}$ turns into 4 sides of $P_{1}$. Generally, one side of $P_{n-1}$ turns into 4 sides of $P_{n}$, so the number of sides of $P_{n}$ is $a_{n}=3 \cdot 4^{n-1}$.

Moreover, $P_{n}$ is generated by adding $a_{n-1}$ small equilateral triangles to $P_{n-1}$, and the side length of each added small equilateral triangle is $\frac{1}{3^{n}}$ of the original equilateral triangle. Therefore, the area of each added small equilateral triangle is $\left(\frac{1}{3^{n}}\right)^{2}$, so
$$
S_{n}=S_{n-1}+a_{n-1} \cdot\left(\frac{1}{3^{n}}\right)^{2}=S_{n-1}+3 \cdot 4^{n-1} \cdot\left(\frac{1}{3^{2 n}}\right)=S_{n-1}+\frac{3}{4}\left(\frac{4}{9}\right)^{n} .
$$

Thus, $S_{n}=S_{0}+\sum_{k=1}^{n}\left(S_{k}-S_{k-1}\right)=1+\sum_{k=1}^{n} \frac{3}{4}\left(\frac{4}{9}\right)^{n}=$
$$
1+\frac{3}{4} \cdot \frac{\frac{4}{9}\left[1-\left(\frac{4}{9}\right)^{n}\right]}{1-\frac{4}{9}}=1+\frac{3}{5}\left[1-\left(\frac{4}{9}\right)^{n}\right]=\frac{8}{5}-\frac{3}{5}\left(\frac{4}{9}\right)^{n} .
$$
(2) $\lim _{n \rightarrow+\infty} S_{n}=\lim _{n \rightarrow+\infty}\left[\frac{8}{5}-\frac{3}{5}\left(\frac{4}{9}\right)^{n}\right]=\frac{8}{5}$.