Let  $f\left(x\right),g\left(x\right):R\to R$  be periodic function with period  $\frac{3}{2}$  and $\frac{1}{2}$ respectively such that ${\text{lim}}_{x\to 0}\frac{f\left(x\right)}{x}=1,{\text{lim}}_{x\to 0}\frac{g\left(x\right)}{x}=2$, then  ${\text{lim}}_{n\to \infty }\frac{f\left(\frac{3}{2}{(3+\sqrt{7})}^n\right)}{g\left(\frac{1}{2}{(2+\sqrt{2})}^n\right)}$ is equal to

$$\begin{gathered} \frac{f\left({(3+\sqrt{7})}^n{T}_1\right)}{g\left({(2+\sqrt{2})}^n{T}_2\right)}=\frac{f({(3+\sqrt{7})}^2{T}_1+{(3-\sqrt{7})}^n{T}_2-{(3-\sqrt{7})}^n{T}_1)}{g({(2+\sqrt{2})}^n{T}_2+{(2-\sqrt{2})}^n{T}_2-{(2-\sqrt{2})}^n{T}_2)} \\ =\frac{f(-{(3-\sqrt{7})}^n{T}_1)}{g(-{(2-\sqrt{2})}^n{T}_2)}\forall n\ge 1 \end{gathered}$$

$$\begin{gathered} {\lim }_{n\to \infty }\frac{f\left(T_1{(3+\sqrt{7})}^n\right)}{f\left(T_2{(2+\sqrt{2})}^n\right)} \\ =\frac{T_1}{T_2}{\lim }_{n\to \infty }\{\frac{f(-{(3-\sqrt{7})}^n{T}_1)}{-{(3-\sqrt{7})}^n{T}_1}.\frac{-{(2-\sqrt{2})}^n{T}_2}{g(-{(2-\sqrt{2})}^n{T}_2)}.\frac{{(3-\sqrt{7})}^n}{{(2-\sqrt{2})}^n}\}=0 \end{gathered}$$