Strange Functional Equation

From $f(x_{1}+x_{2})=f(x_{1})f(x_{2})$ when $x_{1}=0$ we know that $f(0)=1$ It is an even function, and also has x=1 as an asymptote. Therefore, the function must be the same for the period of 2; for example, from x=-1 to x=1 is the same as from x=1 to x=3. Thus, $f(x)=f(x+2n)$ when n is an integer

Hence, $\lim_{n\rightarrow \infty }ln(f(2n+\frac{1}{2n}))=\lim_{n\rightarrow \infty }ln(f(\frac{1}{2n}))=ln(f(0))=ln(1)=0$

It is easy to find out that the function is periodic with $P=2$ , i.e. $f(x)=f(x+2)$ from the given symmetry equations: $f(x)=f(-x)$ and $f(1-x)=f(1+x)$ . Also notice that $f(x_{1}+x_{2})=f(x_{1})\cdot f(x_{2})$ can be extended to $f(x_{1}+x_{2}+x_{3}+...+x_{2n})=f(x_{1})\cdot f(x_{2})\cdot f(x_{3})...f(x_{2n})$ for the same domain. Suppose that $x_{1}=x_{2}=x_{3}=...=x_{2n}=\frac{1}{2n}$ which gives us $f(x_{1}+x_{2}+x_{3}+...+x_{2n})=f(1)=(f(\frac{1}{2n}))^{2n}$ . Then $a_{n}=f(2n+\frac{1}{2n})=f(\frac{1}{2n})=a^\frac{1}{2n}$ . Thus, $\displaystyle\lim_{n\rightarrow\infty}(\ln\:a_{n})=\displaystyle\lim_{n\rightarrow\infty}(\frac{1}{2n}\cdot\ln\:a)=0$