Prove that $\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$ 
Given a sequence $\{a_n\}_{n\in \mathbb N}$ of real numbers such that
$$\begin{align}
\lim_{n\to\infty}\frac{a_na_{n+1} - a_{n-1}a_{n+2}}{a_{n+1}^2 - a_na_{n+2}} &= p + q && (1)\\[1mm]
\lim_{n\to\infty} \frac{a_n^2 - a_{n-1}a_{n+1}}{a_{n+1}^2 - a_na_{n+2}} &= pq && (2)\end{align}$$
where $|p| < |q|$, prove that
$$\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$$


Attempts:
Idea #1: Let us denote $b_n = \dfrac{a_n}{a_{n+1}}$. If we divide both numerators and denominators of $(1)$ and $(2)$ by $a_na_{n+2}$ and $a_{n+1}a_n$, respectively, we have
$$
\begin{align}
\lim_{n\to\infty}\frac{\dfrac{a_{n+1}}{a_{n+2}} - \dfrac{a_{n-1}}{a_{n}}}{\dfrac{a_{n+1}}{a_n}\dfrac{a_{n+1}}{a_{n+2}} - 1} &= p+q && (1') \\[1mm]
\lim_{n\to\infty} \frac{\dfrac{a_n}{a_{a+1}} - \dfrac{a_{n-1}}{a_n}}{\dfrac{a_{n+1}}{a_n} - \dfrac{a_{n+2}}{a_{n+1}}} &= pq && (2')
\end{align}$$
which now can be written as
$$\begin{align}
\lim_{n\to\infty}\frac{b_{n+1}b_n - b_nb_{n-1}}{b_{n+1} - b_n} &= p+q  && (1'') \\[1mm]
\lim_{n\to\infty} \frac{b_n - b_{n-1}}{\dfrac{1}{b_{n+1}} - \dfrac{1}{b_{n}}} &= -pq && (2'')
\end{align}$$
Now, every numerator and denominator contains the difference of consecutive terms of some sequence (reminds me of Cesaro-Stolz, but that cannot be applied here).

Idea #2: The given conditions can be written as
$$\begin{align}
\lim_{n\to\infty}\frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= p + q && (1')\\[2mm]
\lim_{n\to\infty} \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= pq && (2')\end{align}$$
Now, maybe a bit of Linear Algebra could be incorporated somehow.
If we set (B. Grossman)
$$\begin{align} x_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}}, \quad  y_n =  \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} \end{align}$$
Then, we have
$$\pmatrix{a_{n+1} & a_{n+2}\\a_n & a_{n+1}} \pmatrix{x_n\\y_n} = \pmatrix{a_n\\a_{n-1}}$$
This gives
$$\frac{a_{n+1}x_n + a_{n+2}y_n}{a_nx_n + a_{n+1}y_n} = \frac{a_n}{a_{n-1}}$$
Dividing the numerator and the denominator by $a_{n+1}$, we get
$$\frac{x_n + \dfrac{a_{n+2}}{a_{n+1}}y_n}{\dfrac{a_n}{a_{n+1}}x_n + y_n} = \frac{a_n}{a_{n-1}}$$
Noting that $x_n \to p+q$, $y_n \to pq$ and assuming that $\dfrac{a_n}{a_{n-1}} \to A$, and sending $n$ to infinity, we get
$$\frac{p+q + Apq}{\dfrac{1}{A}(p+q) + pq} = A$$
which simplifies to
$$0=0$$
I think I did something wrong somewhere.
Any help is appreciated.
 A: The following proof is not complete, hoped this is helpful to you.
Denote $b_n=\frac{a_n}{a_{n+1}}$ , then
$$\lim\limits_{n\to\infty}\frac{b_n(b_{n+1}-b_{n-1})}{b_{n+1}-b_n}=p+q,\quad \lim\limits_{n\to\infty}\frac{b_n b_{n+1}(b_{n}-b_{n-1})}{b_{n+1}-b_n}=pq.$$
If you assume that $\lim\limits_{n\to\infty} b_n=x$ exists, since
$$p+q=\lim\limits_{n\to\infty}\frac{b_n(b_{n+1}-b_{n-1})}{b_{n+1}-b_n}=\lim\limits_{n\to\infty}[\frac{b_n(b_{n}-b_{n-1})}{b_{n+1}-b_n}+b_n],$$
hence $\lim\limits_{n\to\infty}\frac{b_n(b_{n}-b_{n-1}\ )}{b_{n+1}-b_n}=y$ also exists. Moreover,
$$x+y=p+q,\quad xy=pq\implies
\left\{\begin{array}{l}
x=p \\ y=q 
\end{array}\right.,\ or
\left\{\begin{array}{l}
x=q \\ y=p
\end{array}\right..$$
If you can check that $\lim\limits_{n\to\infty} |b_n|\le \lim\limits_{n\to\infty}|\frac{b_n(b_{n}-b_{n-1}\ )}{b_{n+1}-b_n}|$, then
$$\lim\limits_{n\to\infty} b_n=x=p.$$
A: HINT:
Call
$$
X_n=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n+1}^2-a_na_{n+2}}\\
Y_n=\frac{a_{n}^2-a_{n-1}a_{n+1}}{a_{n+1}^2-a_na_{n+2}}
$$
then
$$
\lim_nX_n=p+q\\
\lim_nY_n=pq
$$
thus
$$
\lim_n\frac{X_n}{Y_n}=\frac{p+q}{pq}
$$
but
$$
\frac{X_n}{Y_n}
=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n}^2-a_{n-1}a_{n+1}}\;.
$$
Then you get
$$
\frac{X_n}{Y_nY_{n-1}}=
\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n-1}^2-a_{n-2}a_{n}}
$$
and inductively
$$
\frac{X_n}{Y_nY_{n-1}\cdots Y_2}=
\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{1}^2-a_{0}a_{2}}
$$
from which
$$
\frac{X_n}{Y_nY_{n-1}\cdots Y_2}(a_{1}^2-a_{0}a_{2})+a_{n-1}a_{n+2}=a_na_{n+1}
$$
divide by $a_{n+1}^2$ and work on LHS.
