10. 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 $x_{n+1}(n=1,2,3, \ldots \ldots)$, when $n$ is sufficiently large, we can use $x_{n}$ as an approximate value of the zero. Let $f(x)=x^{2}-2, x_{1}=2$, find the recursive formula of the sequence $\left\{x_{n}\right\}$, and find its limit.
10. Solution: The equation of the tangent line is $y=f^{\prime}\left(x_{n}\right)\left(x-x_{n}\right)+f\left(x_{n}\right)$. Given $f(x)=x^{2}-2$, the equation of the tangent line is $y=2 x_{n}\left(x-x_{n}\right)+x_{n}^{2}-2=2 x_{n} x-x_{n}^{2}-2$. The intersection point with the $x$-axis is $\left(\frac{x_{n}^{2}+2}{2 x_{n}}, 0\right)$, so the recursive formula is $x_{n+1}=\frac{x_{n}^{2}+2}{2 x_{n}}, x_{1}=2$. Since $x_{1}=2$, and $x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$, by induction we know $\sqrt{2} \leq x_{n} \leq 2$, thus $x_{n+1}-\sqrt{2}=\frac{1}{2 x_{n}}\left(x_{n}-\sqrt{2}\right)^{2}=\frac{x_{n}-\sqrt{2}}{2 x_{n}}\left(x_{n}-\sqrt{2}\right) \leq \frac{1}{2}\left(x_{n}-\sqrt{2}\right)$.

Therefore, $x_{n+1}-\sqrt{2} \leq \frac{1}{2}\left(x_{n}-\sqrt{2}\right) \leq \ldots \leq\left(\frac{1}{2}\right)^{n}\left(x_{1}-\sqrt{2}\right)$. As $n \rightarrow+\infty$, $x_{n} \rightarrow \sqrt{2}$, so the limit of the recursive sequence is $\sqrt{2}$.