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 ${\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}$?

After$$\lim_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{5}}\left(\xi^{n-1}-\phi^{n-1}\right)}{\frac{1}{\sqrt{5}}\left(\xi^{n+1}-\phi^{n+1}\right)},$$you should have obtained$$\lim_{n\to\infty}\frac{\xi^{n-1}\left(1-\frac{\phi^{n-1}}{\xi^{n-1}}\right)}{\xi^{n+1}\left(1-\frac{\phi^{n+1}}{\xi^{n+1}}\right)},$$which is equal to$$\frac1{\xi^2}\lim_{n\to\infty}\frac{1-\frac{\phi^{n-1}}{\xi^{n-1}}}{1-\frac{\phi^{n+1}}{\xi^{n+1}}}=\frac1{\xi^2}.$$
