10. (20 points) In the methods for finding the zeros of a function, there is one called the Newton's method. The process of this method is: for the function $f(x)$, given $x_{1}$, and the intersection of the tangent line of $f(x)$ passing through the point $Q_{n}\left(x_{n}, f\left(x_{n}\right)\right)$ with the $x$-axis is denoted as $x_{n+1}(n=1,2, \cdots)$. As shown in Figure 3, we can use $x_{n}$ as an approximate value of the zero of $f(x)$. Let $f(x)=x^{2}-2$, $x_{1}=2$.
(1) Investigate the recursive formula of the sequence $\left\{x_{n}\right\}$, and provide a proof;
(2) Try to find the limit of the sequence $\left\{x_{n}\right\}$, and explain the meaning of this limit.
(Gold Lei provided the question)
10. (1) Differentiate $f(x)$ to get $f^{\prime}(x)=2 x$. Therefore, the equation of the tangent line passing through $Q\left(x_{n}, f\left(x_{n}\right)\right)$ is
$$
y-f\left(x_{n}\right)=f^{\prime}\left(x_{n}\right)\left(x-x_{n}\right) \Rightarrow y-\left(x_{n}^{2}-2\right)=2 x_{n} \cdot\left(x-x_{n}\right) \text {. }
$$

Setting $y=0$ in the above equation, we get
$$
x=x_{n}-\frac{x_{n}^{2}-2}{2 x_{n}}=\frac{x_{n}^{2}+2}{2 x_{n}} .
$$

Thus, the recursive formula for $\left\{x_{n}\right\}$ is
$$
x_{n+1}=\frac{x_{n}^{2}+2}{2 x_{n}} \quad\left(n \in \mathbf{N}^{*}\right) .
$$
(2) (Method 1) From equation (1), we have
$$
\begin{array}{l}
x_{n+1}+\sqrt{2}=\frac{x_{n}^{2}+2}{2 x_{n}}+\sqrt{2}=\frac{\left(x_{n}+\sqrt{2}\right)^{2}}{2 x_{n}}, \\
x_{n+1}-\sqrt{2}=\frac{x_{n}^{2}+2}{2 x_{n}}-\sqrt{2}=\frac{\left(x_{n}-\sqrt{2}\right)^{2}}{2 x_{n}} .
\end{array}
$$

Dividing the above two equations, we get
$$
\frac{x_{n+1}-\sqrt{2}}{x_{n+1}+\sqrt{2}}=\left(\frac{x_{n}-\sqrt{2}}{x_{n}+\sqrt{2}}\right)^{2} \Rightarrow \lg \frac{x_{n+1}-\sqrt{2}}{x_{n+2}+\sqrt{2}}=2 \lg \frac{x_{n}-\sqrt{2}}{x_{n}+\sqrt{2}} .
$$

Thus, the sequence $\left\{\lg \frac{x_{n}-\sqrt{2}}{x_{n}+\sqrt{2}}\right\}$ is a geometric sequence with the first term $\lg \frac{2-\sqrt{2}}{2+\sqrt{2}}$ and common ratio 2, i.e.,
$$
\lg \frac{x_{n}-\sqrt{2}}{x_{n}+\sqrt{2}}=2^{n-1} \cdot \lg \frac{2-\sqrt{2}}{2+\sqrt{2}} \Rightarrow \frac{x_{n}-\sqrt{2}}{x_{n}+\sqrt{2}}=\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^{2^{n-1}} \text {. }
$$

Therefore,
$$
x_{n}=\sqrt{2} \cdot \frac{(2+\sqrt{2})^{2^{n-1}}+(2-\sqrt{2})^{2^{n-1}}}{(2+\sqrt{2})^{2^{n-1}}-(2-\sqrt{2})^{2^{n-1}}} .
$$

Thus,
$$
\begin{aligned}
\lim _{n \rightarrow \infty} x_{n} & =\sqrt{2} \cdot \lim _{n \rightarrow \infty} \frac{(2+\sqrt{2})^{2^{n-1}}+(2-\sqrt{2})^{2^{n-1}}}{(2+\sqrt{2})^{2^{n-1}}-(2-\sqrt{2})^{2^{n-1}}} \\
& =\sqrt{2} \cdot \lim _{n \rightarrow \infty} \frac{1+(3-2 \sqrt{2})^{2^{n-1}}}{1-(3-2 \sqrt{2})^{2^{n-1}}}=\sqrt{2} .
\end{aligned}
$$

In fact, $\sqrt{2}$ is the zero of $f(x)$. This indicates that when this method is performed for a sufficient number of steps, the approximate value of the zero of $f(x)$ obtained is very close to the actual value.
(Method 2) Below is a method to find the limit of $\left\{x_{n}\right\}$ without needing the general term.
We claim that for any positive integer $n$, we have $x_{n}>x_{n+1}>\sqrt{2}$. We will prove this by mathematical induction.
From $x_{1}=2$ and equation (1), we can calculate $x_{2}=\frac{3}{2}$, and it is easy to see that $x_{1}>x_{2}>\sqrt{2}$. Thus, the claim holds for $n=1$.
Assume that the claim holds for $n=k$, i.e., $x_{k}>x_{k+1}>\sqrt{2}$. Then
$$
x_{k+2}=\frac{x_{k+1}^{2}+2}{2 x_{k+1}} \geqslant \frac{2 \sqrt{2 \cdot x_{k+1}^{2}}}{2 x_{k+1}}=\sqrt{2} \text {, }
$$

with equality if and only if $x_{k+1}=\sqrt{2}$, which is impossible, so $x_{k+2}>\sqrt{2}$. Also,
$$
x_{k+2}-x_{k+1}=-\frac{\left(x_{k+1}-\sqrt{2}\right)\left(x_{k+1}+\sqrt{2}\right)}{2 x_{k+1}}x_{k+2}$. Thus, $x_{k+1}>x_{k+2}>\sqrt{2}$, i.e., the claim also holds for $n=k+1$.
Therefore, for any positive integer $n$, we have $x_{n}>x_{n+1}>\sqrt{2}$. This indicates that the sequence $\left\{x_{n}\right\}$ is monotonically decreasing and bounded below, so $\lim _{n \rightarrow \infty} x_{n}$ exists. Let $\lim _{n \rightarrow \infty} x_{n}=A$, and take the limit on both sides of equation (1), we get
$$
A=\frac{A^{2}+2}{2 A} \text {. }
$$

Solving this, we get $A=\sqrt{2}$ (discarding the negative value), i.e., $\lim _{n \rightarrow \infty} x_{n}=\sqrt{2}$.