Example 1 Given the parabola $y=x^{2}$, draw a line with a slope of 1 through the origin intersecting the parabola at $P_{1}$, draw a line with a slope of $\frac{1}{2}$ through $P_{1}$ intersecting the parabola at $P_{2}, \cdots \cdots$, generally, draw a line with a slope of $2^{-n}$ through the point $P_{n}$ intersecting the parabola at $P_{n+1}$, let the coordinates of $P$ be $\left(x_{n}, y_{n}\right)$. Find the limit point of $P_{2 n+1}$.
Given the conditions, we have
$$
\left\{\begin{array}{l}
\frac{y_{n+1}-y_{n}}{x_{n+1}-x_{n}}=2^{-n}, \\
y_{n}=x_{n}^{2} .
\end{array}\right.
$$
Substituting equation (2) into equation (1) and rearranging, we get
$$
x_{n+1}+x_{n}=2^{-n} \text {, }
$$

which can be rewritten as
$$
(-1)^{n+1} x_{n+1}=(-1)^{n} x_{n}-\left(-\frac{1}{2}\right)^{n} \text {. }
$$

Summing both sides with respect to $n$, we obtain
$$
\begin{array}{l}
(-1)^{n+1} x_{n+1}=-x_{1}+\frac{1-\left(-\frac{1}{2}\right)^{n}}{3} . \\
\therefore x_{n}=\frac{\left(\frac{1}{2}\right)^{n-1}+(-1)^{n-1} 3 x_{1}+(-1)^{n}}{3} .
\end{array}
$$

Furthermore,
$$
\begin{array}{l}
y_{n}=x_{n}^{2} \\
=\left[\frac{\left(\frac{1}{2}\right)^{n-1}+(-1)^{n-1} 3 x_{1}+(-1)^{n}}{3}\right]^{2}, \\
x_{2 n+1}=\frac{\left(\frac{1}{2}\right)^{2 n}+3 x_{1}-1}{3}, \\
y_{2 n+1}=\left[\frac{\left(\frac{1}{2}\right)^{2 n}+3 x_{1}-1}{3}\right]^{2} .
\end{array}
$$

As $n \rightarrow \infty$,
$$
x_{2 n+1} \rightarrow \frac{3 x_{1}-1}{3}, y_{2 n+1} \rightarrow\left(\frac{3 x_{1}-1}{3}\right)^{2} \text {. }
$$

Solving $\left\{\begin{array}{l}y=x, \\ y=x^{2}\end{array}\right.$, we get $x_{1}=1$.
$\therefore$ The limit point of $P_{2 n+1}$ is $+\left(\frac{2}{3}, \frac{4}{9}\right)$.