14. Let $f(x)$ 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 $f\left(\frac{1}{2}\right)$ and $f\left(\frac{1}{4}\right)$;
(2) Prove that $f(x)$ is a periodic function;
(3) Let $a_{n}=f\left(2 n+\frac{1}{2 n}\right)$, find $\lim _{n \rightarrow \infty}\left(\ln a_{n}\right)$.
14. (1) $f(x)=f\left(\frac{x}{2}\right) \cdot f\left(\frac{x}{2}\right) \geqslant 0, f(1)=\left[f\left(\frac{1}{2}\right)\right]^{2}, f\left(\frac{1}{2}\right)=\left(f\left(\frac{1}{4}\right)\right)^{2}, f(1)=a>0$, so $f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}, f\left(\frac{1}{4}\right)=a^{\frac{1}{4}}$
(2) $f(x)=f(2-x)$, and $f(-x)=f(x)$, so $f(-x)=f(2-x), f(x)=f(2+x)$, hence $f(x)$ is a periodic function on $\mathbf{R}$, and 2 is one of its periods.
(3) $f\left(\frac{1}{2}\right)=f\left(n \cdot \frac{1}{2 n}\right)=f\left(\frac{1}{2 n}+(n-1) \cdot \frac{1}{2 n}\right)=f\left(\frac{1}{2 n}\right) \cdot\left(f(n-1) \cdot \frac{1}{2 n}\right)=\cdots=f\left(\frac{1}{2 n}\right) \cdot$ $f\left(\frac{1}{2 n}\right) \cdots f\left(\frac{1}{2 n}\right)=\left[f\left(\frac{1}{2 n}\right)\right]^{n}, f\left(\frac{1}{2}\right)=a^{\frac{1}{2}}$, so $f\left(\frac{1}{2 n}\right)=a^{\frac{1}{2 n}}$, and the period of $f(x)$ is 2. $f\left(\frac{1}{2 n}+2 n\right)=$ $f\left(\frac{1}{2 n}\right), a_{n}=a^{\frac{1}{2 n}}$, therefore, $\lim _{n \rightarrow \infty}\left(\ln a_{n}\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{2 n} \ln a\right)=0$.