2. Solution: Let the side length of the equilateral $\triangle ABC$ be $a, BP_{n}=x_{n}$, then $BP_{n+1}=x_{n+1}$
$$
\begin{array}{l}
B Q_{n}=B P_{n} \cos 60^{\circ}=\frac{1}{2} x_{n}, Q_{n} C=a-\frac{1}{2} x_{n} \\
C R_{n}=C Q_{n} \cos 60^{\circ}=\frac{1}{2}\left(a-\frac{x_{n}}{2}\right) \\
A R_{n}=a-C R_{n}=\frac{1}{2} a+\frac{1}{4} x_{n} \\
A P_{n+1}=A P_{n} \cos 60^{\circ}=\frac{1}{2} A P_{n}=\frac{1}{4} a+\frac{1}{8} x_{n} \\
B P_{n+1}=x_{n+1}=a-\frac{1}{4} a-\frac{1}{8} x_{n}=\frac{3 a}{4}-\frac{1}{8} x_{n} .
\end{array}
$$

Thus, $\quad x_{n+1}=-\frac{1}{8} x_{n}+\frac{3}{4} a$
So, $x_{n+1}-\frac{2}{3} a=-\frac{1}{8}\left(x_{n}-\frac{2}{3} a\right)$
That is, the sequence $\left\{x_{n}-\frac{2 a}{3}\right\}$ is a geometric sequence with a common ratio of $-\frac{1}{8}$. Therefore,
$$
x_{n}-\frac{2 a}{3}=\left(x_{1}-\frac{2 a}{3}\right)\left(-\frac{1}{8}\right)^{n-1}, \lim _{n \rightarrow \infty} x_{n}=\frac{2 a}{3} .
$$
2. Let $P_{1}$ be a point on the side $A B$ of the equilateral $\triangle A B C$, draw a perpendicular from $P_{1}$ to $B C$, with the foot of the perpendicular being $Q_{1}$, draw a perpendicular from $Q_{1}$ to $C A$, with the foot of the perpendicular being $R_{1}$, draw a perpendicular from $R_{1}$ to $A B$, with the foot of the perpendicular being $P_{2}$, and then repeat the same process from $P_{2}$, obtaining $Q_{2}$, $R_{2}$, $P_{3}$, $Q_{3}$, $R_{3}$, $\cdots$, as $n \rightarrow \infty$, to which point does $P_{n}$ tend?