Find $\lim\limits_{n\to\infty}\frac{x_{n+1}}{x_n}! $ where $x_n=x_{n-1}+x_{n-2} ,(n>2),x_1=1,x_2=2$
$x_n=x_{n-1}+x_{n-2}$
$x_{n+1}=x_{n}+x_{n-1}$
From the first recurrence relation, $$x_n=\frac{3+\sqrt{5}}{5+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n+\frac{3-\sqrt{5}}{5-\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$$
From the second recurrence relation,
$$x_{n+1}=c_2({x_1}^{'})^{n+1}+c_3({x_2}^{'})^{n+1}$$
where $${x_2}^{'}=\frac{1+\sqrt{5}}{2},{x_3}^{'}=\frac{1-\sqrt{5}}{2}$$ $$c_2=\frac{x_3-{x_3}^{'}x_2}{{x_2}^{'}({x_2}^{'}-{x_3}^{'})}=\frac{2(2+\sqrt{5})}{5+\sqrt{5}}$$ $$c_3=-\frac{x_3-{x_2}^{'}x_2}{{x_3}^{'}({x_2}^{'}-{x_3}^{'})}=\frac{2(2-\sqrt{5})}{5-\sqrt{5}}$$
gives $$x_{n+1}=\frac{2(2+\sqrt{5})}{5+\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}+\frac{2(2-\sqrt{5})}{5-\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}$$
This gives $$\lim\limits_{n\to\infty}\frac{x_{n+1}}{x_n}! =\frac{2(2+\sqrt{5})}{3+\sqrt{5}}!$$
Is this correct?