Recurrence relation/with limit Let $F_{n+1}=F_{n-1}+F_{n-2}$ for $n \in \mathbb{N}$ with $n \geq 2$
$F_0:=0$ and $F_1:=1$.
How to compute
$\lim\limits_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}$?
I tried to use Binet's formula:
$\lim\limits_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}=\lim\limits_{n\to\infty}\frac{\frac{1}{\sqrt{5}}(\xi^{n-1}-\phi^{n-1})}{\frac{1}{\sqrt{5}}(\xi^{n+1}-\phi^{n+1})}=\lim\limits_{x\to\infty}\frac{\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}{\xi^{n+1}(1-\frac{\phi^{n+1}}{\xi^{n+1}})}$
But I don't know what to do next. 
I suppose $\xi^{n+1}(1-\frac{\phi^{n+1}}{\xi^{n+1}})=\xi$ but what about ${\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}$?

We have$$\lim\limits_{n\to\infty}\frac{\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}{\xi^{n+1}(1-\frac{\phi^{n+1}}{\xi^{n+1}})}={1\over \xi ^2}\lim_{n\to \infty}{1-\left({\phi\over \xi}\right)^{n-1}\over 1-\left({\phi\over \xi}\right)^{n+1}}$$where $${\phi\over \xi}={{1-\sqrt 5\over 2}\over {1+\sqrt 5\over 2}}={1-\sqrt 5\over 1+\sqrt 5}$$therefore $-1<{\phi \over \xi}<0$and we obtain $$\lim_{n\to \infty}{1-\left({\phi\over \xi}\right)^{n-1}\over 1-\left({\phi\over \xi}\right)^{n+1}}={1\over \xi ^2}={3-\sqrt 5\over 2}$$
