How prove this limit $\lim\limits_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a$ given two other limits related to $f_n$ 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 

I'm going to post a failed approach, just so you can see what I tried.  This is here only to show an approach and (maybe) generate new ideas.
Let $$x:=a+b\;\;\text{ and }\;\; y:=ab.$$  Assume for now $0<a<b$.  Then 
$$\frac{x-\sqrt{x^2-4y}}{2}
=\frac{a+b-\sqrt{(a+b)^2-4ab}}{2}
=\frac{a+b-\sqrt{(a-b)^2}}{2}
=\frac{a+b-\vert a-b\vert}{2}
=\frac{a+b-b+a}{2}
=a.$$
Now we replace 
$a+b$ with $\lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}$ and 
$ab$ with  $\lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}$, and due to continuity move the limit outside:
$$\frac{a+b-\sqrt{(a+b)^2-4ab}}{2}
=\lim_{n\to\infty}
\frac{\left(\frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}\right)-\sqrt{\left(\frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}\right)^2
-4\left(\frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}\right)}}{2}
=\lim_{n\to\infty}
\frac{\left({f_{n+1}f_n-f_{n-1}f_{n+2}}\right)
-\sqrt{\left({f_{n+1}f_n-f_{n-1}f_{n+2}}\right)^2
-4\left({f_{n}^2-f_{n-1}f_{n+1}}\right)\left({f_{n+1}^2-f_nf_{n+2}}\right)}}{2\left({f_{n+1}^2-f_nf_{n+2}}\right)}
=\lim_{n\to\infty}
\frac{f_{n+1}f_n-f_{n-1}f_{n+2}
-\sqrt{
f_{n+1}^2f_n^2
-2f_{n+1}f_n
f_{n-1}f_{n+2}
+f_{n-1}^2f_{n+2}^2
-4f_n^2f_{n+1}^2+4f_{n}^2f_nf_{n+2}+4f_{n-1}f_{n+1}f_{n+1}^2-4f_{n-1}f_{n+1}f_nf_{n+2}}
}{2\left({f_{n+1}^2-f_nf_{n+2}}\right)}
$$
At this point there are no useful simplifications to be done inside the radical.  It does not seem to agree with the $x$ and $y$ case.
