Example 17 Let the infinite sequence be defined by $x_{1}$ and $x_{n}=\sqrt{2 x_{n-1}+3}$, find $\lim _{n \rightarrow \infty} x_{n}$.
Solving for the fixed point of $f(x)$ yields $x_{0}=3\left(x_{0} \geqslant 0\right)$. The fixed point $x_{0}=3$ suggests making the substitution $|x_{0}-3|$, we have
$$
0<\left|x_{n}-3\right|=\frac{2\left|x_{n-1}-3\right|}{\sqrt{2 x_{n-1}-3}+3} \leqslant \frac{2}{3}\left|x_{n-1}-3\right| \leqslant \cdots \leqslant\left(\frac{2}{3}\right)^{n-1} \cdot\left|x_{1}-3\right| \text {, }
$$

and $\lim _{n \rightarrow \infty}\left(\frac{2}{3}\right)^{n-1}\left|x_{1}-3\right|=0$, then $\lim _{n \rightarrow \infty}\left|x_{n}-3\right|=0$, hence $\lim _{n \rightarrow \infty} x_{n}=3$.