TITLE: How prove this limit $\lim\limits_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a$ given two other limits related to $f_n$
QUESTION [7 upvotes]: Let $(f_n)$- real sequence such that $$ \lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}=a+b, $$ and $$ \lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}=ab   \quad    (|a|<|b|). $$
Prove that:$$\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a $$
I think we must prove $\displaystyle\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}} $ exists,and we prove this limit is $a$,But I can't prove this limit exists.
My idea: since
$$\lim_{n\to\infty}\dfrac{\dfrac{f_{n}}{f_{n+1}}-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=a+b$$
and 
$$\lim_{n\to\infty}\dfrac{\left(\dfrac{f_{n}}{f_{n+1}}\right)^2-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=ab$$
But I felt this deal is not useful,
Other idea: I want to take the 
Fibonacci sequence  to solve this problem, But I can't, 
Thank you

REPLY [5 votes]: This is a supplement to several other answers about existence of the limit.
Let $P_n = \left|\begin{matrix}f_{n+1} & f_{n+2}\\ f_{n-1} & f_n\end{matrix}\right|$
and $Q_n = \left|\begin{matrix}f_{n} & f_{n+1} \\ f_{n-1} & f_n\end{matrix}\right|$,
the given condition can be rewritten as
$$\lim_{n\to\infty} \frac{P_n}{Q_{n+1}} = a + b
\quad\text{ and }\quad\lim_{n\to\infty} \frac{Q_n}{Q_{n+1}} =  ab\tag{*1}$$
Notice 
$$f_n P_n - f_{n+1}Q_n = f_{n-1} Q_{n+1}
\quad\iff\quad f_{n+1} = \frac{P_n}{Q_n} f_n - \frac{Q_{n+1}}{Q_n} f_{n-1}
$$
The sequence satisfies a non-linear recurrence relations whose coefficients converge to
some finite limit. The "limit" of the recurrence relation has the form
$$f_{n+1} = \left(\frac{1}{a} + \frac{1}{b}\right) f_{n} - \frac{1}{ab} f_{n-1}$$
with characteristic polynomial $\lambda^2 - \left(\frac{1}{a} + \frac{1}{b}\right)\lambda + \frac{1}{ab} = \left(\lambda - \frac{1}{a}\right)\left(\lambda - \frac{1}{b}\right)$.
By assumption, $|a| < |b|$. This means distinct roots of this characteristic polynomial has distinct modulus. By Poincare-Perron theorem$\color{blue}{^{[1]}}$, we have either

$f_n = 0$ for all sufficiently large $n$ or
$\lim\limits_{n\to\infty} \frac{f_{n+1}}{f_n}$ exists and equal to $\frac{1}{a}$ or $\frac{1}{b}$.

The assumption $(*1)$ is written in such a way that $Q_{n+1}$ is non-zero for sufficient large $n$. So we can ignore the possibility that $f_n$ vanishes for large $n$. This implies the limit $\lim\limits_{n\to\infty} \frac{f_{n}}{f_{n+1}}$ does exist and equal to either $a$ or $b$.
Notes

$\color{blue}{[1]}$ For more details of Poincare-Perron theorem and a proof of it, please consult Chapter 8 of Saber Elaydi's book An Introduction to Difference Equations.