Example 1 Given a parabola $y=x^{2}$, draw a line with a slope of 1 through the origin $O$ intersecting the parabola at point $P_{1}$, draw a line with a slope of $\frac{1}{2}$ through point $P_{1}$ intersecting the parabola at point $P_{2}$, then draw a line with a slope of $\frac{1}{4}$ through point $P_{2}$ intersecting the parabola at point $P_{3}, \cdots$, and draw a line with a slope of $\frac{1}{2^{n}}$ through point $P_{n}$ intersecting the parabola at point $P_{n+1}, \cdots$. Let the coordinates of point $P_{n}$ be $\left(x_{n}, y_{n}\right)$, try to find $x_{2 n+1}$, and determine the point to which $P_{1} 、 P_{3}, P_{5}, \cdots$ tend.
Solution: As shown in Figure 4-1, the slope of the line $P_{2 n} P_{2 n-1}$ is
$\frac{x_{2 n+1}^{2}-x_{2 n}^{2}}{x_{2 n+1}-x_{2 n}}=x_{2 n+1}+x_{2 n}$.
Thus, $x_{2 n+1}+x_{2 n}=\frac{1}{2^{2 n}}$.
Similarly, we get
$x_{2 n}+x_{2 n-1}=\frac{1}{2^{2 n-1}}$, i.e., $-x_{2 n}-x_{2 n-1}=-\frac{1}{2^{n-1}}$.
Obviously, $x_{1}=1$.
Let $n=1,2,3, \cdots$, we have
$-x_{2}-x_{1}=-\frac{1}{2}$,
$x_{3}+x_{2}=\frac{1}{2^{2}}$,
$\qquad$
$x_{2 n-1}+x_{2(n-1)}=\frac{1}{2^{2(n-1)}}$,
$-x_{2 n}-x_{2 n-1}=-\frac{1}{2^{2 n-1}}$,
$x_{2 n+1}+x_{2 n}=\frac{1}{2^{2 n}}$.
Adding all the above equations, we get
$$
\begin{aligned}
\text { Left side } & =x_{1}-x_{2}-x_{1}+x_{3}+x_{2}-\cdots+x_{2 n-1}+x_{2 n-2}-x_{2 n}-x_{2 n-1}+x_{2 n+1}+x_{2 n} \\
& =x_{2 n+1}, \\
\text { Right side } & =1-\frac{1}{2}+\frac{1}{2^{2}}-\cdots+\frac{1}{2^{2 n-2}}-\frac{1}{2^{2 n-1}}+\frac{1}{2^{2 n}} \\
& =1-\left(\frac{1}{2}-\frac{1}{2^{2}}+\cdots-\frac{1}{2^{2 n-2}}+\frac{1}{2^{2 n-1}}-\frac{1}{2^{2 n}}\right) \\
& =1-\frac{\frac{1}{2}\left[1-\left(-\frac{1}{2}\right)^{2 n}\right]}{1-\left(-\frac{1}{2}\right)}=1-\frac{1}{3}\left[1-\left(\frac{1}{2}\right)^{2 n}\right] \\
& =\frac{2}{3}+\frac{1}{3} \times 4^{-n}=\frac{1}{3}\left(2+4^{-n}\right),
\end{aligned}
$$

Right side $=1-\frac{1}{2}+\frac{1}{2^{2}}-\cdots+\frac{1}{2^{2 n-2}}-\frac{1}{2^{2 n-1}}+\frac{1}{2^{2 n}}$

Therefore, $x_{2 n+1}=\frac{1}{3}\left(2+4^{-n}\right)$.
$$
\lim _{n \rightarrow \infty} x_{2 n+1}=\lim _{n \rightarrow \infty} \frac{1}{3}\left(2+4^{-n}\right)=\frac{2}{3} \text {. }
$$

Thus, the points $P_{1}, P_{3}, P_{5}, \cdots$ approach the point $\left(\frac{2}{3}, \frac{4}{9}\right)$.