Example 21 (2001 National College Entrance Examination Question) Let the function $y=f(x)$ be an even function defined on $\mathbf{R}$, whose graph is symmetric about the line $x=1$. For any $x_{1}, x_{2} \in\left[0, \frac{1}{2}\right]$, we have $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right)$, and $f(1)=a>0$.
(1) Find $f\left(\frac{1}{2}\right) 、 f\left(\frac{1}{4}\right)$;
(2) Prove that $f(x)$ is a periodic function;
(3) Let $a_{n}=f\left(2 n+\frac{1}{2 n}\right)$, find $\lim _{n \rightarrow \infty}\left(\ln a_{n}\right)$.
From $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right) \cdot f\left(x_{2}\right)$, we can think of the exponential function model $y=a^{x}\left(0 \leqslant x \leqslant \frac{1}{2}\right)$ to solve the problem.
(1) $f(x)=f\left(\frac{x}{2}+\frac{x}{2}\right)=f^{2}\left(\frac{x}{2}\right) \geqslant 0, x \in[0,1]$,
and $a=f(1)=f\left(\frac{1}{2}+\frac{1}{2}\right)=f^{2}\left(\frac{1}{2}\right)>0$,
thus $f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}, f\left(\frac{1}{4}\right)=\left[f\left(\frac{1}{2}\right)\right]^{\frac{1}{2}}=a^{\frac{1}{4}}$.
(2) According to the problem, $f(x)=f(-x)$ and $f(1-x)=f(1+x)$, by conclusion 5(1), we know that $f(x)$ is a periodic function, and 2 is one of its positive periods.
(3) From (1), we know $f(x) \geqslant 0, x \in[0,1]$.
Also, $f\left(\frac{1}{2}\right)=f\left(n \cdot \frac{1}{2 n}\right)=f\left[\frac{1}{2 n}+(n-1) \cdot \frac{1}{2 n}\right]$
$=f\left(\frac{1}{2 n}\right) \cdot f\left[(n-1) \cdot \frac{1}{2 n}\right]=\cdots=f\left(\frac{1}{2 n}\right) \cdot f\left(\frac{1}{2 n}\right) \cdots f\left(\frac{1}{2 n}\right)=\left[f\left(\frac{1}{2 n}\right)\right]^{n}$,
and $f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}, f\left(\frac{1}{2 n}\right)=a^{\frac{1}{2 n}},$ and the period of $f(x)$ is 2, so $f\left(2 n+\frac{1}{2 n}\right)=f\left(\frac{1}{2 n}\right)$, which means $a_{n}=a^{\frac{1}{2}}$.
Therefore, $\lim _{n \rightarrow \infty}\left(\ln a_{n}\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{2 n} \ln a\right)=0$.