prove existence of the limit of a sequence So I have the following problem:
$ x_0 = 1 , x_1 = 2 , $and $x_{n+1} = 2x_n + x_{n-1} $for $ n \geq 1.$ 

Show that: $\hspace{2mm} \lim_{n\to \infty} \frac{x_n}{x_{n+1}}   $ exists. 

Then show that the Limit is equal to $\sqrt{2}-1$.

For this I thought i could use the fact that $x_n$ is bounded and I thought that it was monotonically falling, but that is not the case, so I ran out of ideas. And I don't know how to calculate the limit... Thank you very much for your help!
Without guessing the limit you may proceed as follows:


Set $q_n = \frac{x_n}{x_{n+1}}$
$$\Rightarrow x_{n+1} = 2x_n + x_{n-1} \Leftrightarrow \frac{1}{q_n} = 2+q_{n-1}\Leftrightarrow q_n = \frac{1}{2+q_{n-1}}$$


Now, it follows
$$|q_{n+1} - q_n| = \left|\frac{1}{2+q_{n}} - \frac{1}{2+q_{n-1}} \right| = \left|\frac{q_{n-1} - q_n}{(\color{blue}{2}+q_{n})(\color{blue}{2}+q_{n-1})}\right|$$ $$&lt; \frac{1}{\color{blue}{2\cdot 2}}\left|q_{n-1} - q_{n} \right| \stackrel{\mbox{see below}}{\Rightarrow} \boxed{(q_n) \mbox{ is convergent}}$$

So, we get for the limit
$$L =\frac{1}{2+L} \Leftrightarrow (L+1)^2=2 \stackrel{L&gt;0}{\Rightarrow}\boxed{L = \sqrt{2}-1}$$

Edit after comment:

Additional info concerning the convergence of the sequence:


$q_{n+1} = q_1 + \sum_{k=1}^n(q_{k+1} -q_k)$
The sums converge (absolutely) as $|q_{k+1} -q_k| &lt; \left( \frac{1}{4}\right)^{k-1}|q_2 - q_1|$ since
$\left|\sum_{k=1}^{\infty}(q_{k+1} -q_k) \right| \leq \sum_{k=1}^{\infty}|q_{k+1} -q_k| &lt; |q_2 - q_1|\sum_{k=1}^{\infty} \left( \frac{1}{4}\right)^{k-1} &lt; \infty$
As $s_n = \sum_{k=1}^n(q_{k+1} -q_k)$ converges, it follows that $q_{n+1} =q_1 + s_n$ converges
Let

$$\lim_{n\to \infty} \frac{x_n}{x_{n+1}} = \lim_{n\to \infty} \frac{x_{n-1}}{x_{n1}} = k$$

Now

$$\lim_{n\to \infty} \frac{x_n}{x_{n+1}} = k$$
$$\lim_{n\to \infty} \frac{x_n}{2x_{n} + x_{n-1}} = k$$

Take $x_n$ out from numerator and denominator
$$\lim_{n\to \infty} \frac{1}{2 + \frac{x_{n-1}}{x_n}} = k$$

Using the first equation

$$ \frac{1}{2 + k} = k$$
$$k^2+2k-1=0$$
This gives two solutions $k=\sqrt{2}-1$ and $k=-\sqrt{2}-1$. Since none of the terms can be negative, we reject . the second solution thereby giving us

$$\lim_{n\to \infty} \frac{x_n}{x_{n+1}} = k = \sqrt2 - 1$$

EDIT - As suggested by the commenter we need to prove that it is a finite limit before we start with the proof. Initially it's a $\frac{\infty}{\infty}$ form as both $x_n$ and $x_{n+1}$ approach $\infty$ as $n$ approaches $\infty$. I'll use a finite upper bound to show that the limit is finite which means it exists.

For any $n$

$$\frac{x_n}{x_{n+1}} =\frac{x_n}{2x_n + x_{n-1}}$$

As $x_{n-1}$ is always a positive quantity
$$\frac{x_n}{x_{n+1}} \leq \frac{x_n}{2x_n}$$
$$\frac{x_n}{x_{n+1}} \leq \frac{1}{2}$$

For any $n$, you can take the last statement to prove the monotonicity as
$$x_{n+1}\geq x_n$$
And since both $x_n$ and $x_{n+1}$ are positive values, the lower bound is $0$. The upper bound along with lower bound and the monotonicity proves that the limit is finite.