Five. (Total 20 points) As shown in the figure,
$3, \triangle ABC$ is an isosceles triangle with the vertex angle $\frac{\pi}{5}$,
denoted as $\triangle_{1}$, and its vertex angle is denoted as $\alpha_{1}$ $\left(\alpha_{1}=\angle BAC\right)$. Using one of the base angles of $\triangle_{1}$ $(\angle ABC)$ as the vertex angle, construct an isosceles triangle $\triangle_{2}$ (i.e., $\triangle BCD$), with its vertex angle denoted as $\alpha_{2}\left(\alpha_{2}=\angle DBC\right)$; then, using one of the base angles of $\triangle_{2}$ $(\angle BCD)$ as the vertex angle, construct an isosceles triangle $\triangle_{3}$ (i.e., $\triangle CEB$), with its vertex angle denoted as $\alpha_{3}\left(\alpha_{3}=\angle BCE\right)$. Continue this process to construct isosceles triangles $\triangle_{1}, \triangle_{2}, \cdots, \triangle_{n}$, $\cdots$, and the corresponding vertex angles form the sequence $\left\{\alpha_{n}\right\}$.
(1) Find $\alpha_{2}, \alpha_{3}, \alpha_{4}$;
(2) Find the general term formula for $\alpha_{n}$;
(3) Find $\lim _{n \rightarrow \infty} \alpha_{n}$.
(1) According to the problem, given $\alpha_{1}=\frac{\pi}{5}$, we have
$$
\begin{array}{l}
\alpha_{2}=\frac{1}{2}\left(\pi-\alpha_{1}\right)=\frac{1}{2}\left(\pi-\frac{\pi}{5}\right)=\frac{2 \pi}{5}, \\
\alpha_{3}=\frac{1}{2}\left(\pi-\alpha_{2}\right)=\frac{1}{2}\left(\pi-\frac{2 \pi}{5}\right)=\frac{3 \pi}{10}, \\
\alpha_{4}=\frac{1}{2}\left(\pi-\alpha_{3}\right)=\frac{1}{2}\left(\pi-\frac{3 \pi}{10}\right)=\frac{7 \pi}{20} .
\end{array}
$$
(2) By the sum of the interior angles of a triangle, we have
$$
\alpha_{n}=\frac{1}{2}\left(\pi-\alpha_{n-1}\right), \alpha_{n+1}=\frac{1}{2}\left(\pi-\alpha_{n}\right) \text {. }
$$

Subtracting these, we get $\alpha_{n+1}-\alpha_{n}=-\frac{1}{2}\left(\alpha_{n}-\alpha_{n-1}\right)$.
This indicates that the sequence $\left\{\alpha_{n+1}-\alpha_{n}\right\}$ is a geometric sequence with the first term $\alpha_{2}-\alpha_{1}=\frac{\pi}{5}$ and common ratio $q=-\frac{1}{2}$, so we have
$$
\begin{array}{l}
\alpha_{n+1}-\alpha_{n}=\left(\alpha_{2}-\alpha_{1}\right)\left(-\frac{1}{2}\right)^{n-1} \\
=\left(-\frac{1}{2}\right)^{n-1} \cdot \frac{\pi}{5} .
\end{array}
$$

Thus,
$$
\begin{aligned}
\alpha_{n} & =\left(\alpha_{n}-\alpha_{n-1}\right)+\cdots+\left(\alpha_{2}-\alpha_{1}\right)+\alpha_{1} \\
& =\left(\left(-\frac{1}{2}\right)^{n \cdot 2}+\cdots+\left(-\frac{1}{2}\right)+1\right) \frac{\pi}{5}+\frac{\pi}{5} \\
& =\frac{\pi}{3}+\left(-\frac{1}{2}\right)^{n} \cdot \frac{4 \pi}{15} .
\end{aligned}
$$

This is the general term of the sequence.
(3) Since $\lim _{n \rightarrow \infty}\left(-\frac{1}{2}\right)^{n}=0$, we have
$$
\lim _{n \rightarrow \infty}\left(-\frac{1}{2}\right)^{n} \cdot \frac{4 \pi}{15}=\frac{4 \pi}{15} \lim _{n \rightarrow \infty}\left(-\frac{1}{2}\right)^{n}=0,
$$

Thus, $\lim _{n \rightarrow \infty} \alpha_{n}=\lim _{n \rightarrow \infty} \frac{\pi}{3}+\lim _{n \rightarrow \infty}\left(-\frac{1}{2}\right)^{n} \cdot \frac{4 \pi}{15}=\frac{\pi}{3}$.