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!
 A: 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
$$|q_{n+1} - q_n| = \left|\frac{1}{2+q_{n}} - \frac{1}{2+q_{n-1}} \right| = \left|\frac{q_{n-1} - q_n}{(\color{blue}{2}+q_{n})(\color{blue}{2}+q_{n-1})}\right|$$ $$< \frac{1}{\color{blue}{2\cdot 2}}\left|q_{n-1} - q_{n} \right| \stackrel{\mbox{see below}}{\Rightarrow} \boxed{(q_n) \mbox{ is convergent}}$$
So, we get for the limit
$$L =\frac{1}{2+L} \Leftrightarrow (L+1)^2=2 \stackrel{L>0}{\Rightarrow}\boxed{L = \sqrt{2}-1}$$
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 $|q_{k+1} -q_k| < \left( \frac{1}{4}\right)^{k-1}|q_2 - q_1|$ since

*$\left|\sum_{k=1}^{\infty}(q_{k+1} -q_k) \right| \leq \sum_{k=1}^{\infty}|q_{k+1} -q_k| < |q_2 - q_1|\sum_{k=1}^{\infty} \left( \frac{1}{4}\right)^{k-1} < \infty$

*As $s_n = \sum_{k=1}^n(q_{k+1} -q_k)$ converges, it follows that $q_{n+1} =q_1 + s_n$ converges

A: We have the general term. With it, the existence and the value for the limit are proven.
For some values of $A$ and $B$ we have.
$$x_n=A(1-\sqrt{2})^n+B(1+\sqrt{2})^n$$
$$L=\lim_{n\to+\infty}\dfrac{x_n}{x_{n+1}}=\lim_{n\to+\infty}\dfrac{A(1-\sqrt{2})^n+B(1+\sqrt{2})^n}{A(1-\sqrt{2})^{n+1}+B(1+\sqrt{2})^{n+1}}$$
$$L=\lim_{n\to+\infty}\dfrac{\dfrac{A(1-\sqrt{2})^n}{(1+\sqrt{2})^n}+\dfrac{B(1+\sqrt{2})^n}{(1+\sqrt{2})^n}}{\dfrac{A(1-\sqrt{2})^{n+1}}{(1+\sqrt{2})^n}+\dfrac{B(1+\sqrt{2})^{n+1}}{(1+\sqrt{2})^n}}$$
$$L=\dfrac{0+B}{0+B(1+\sqrt{2})}=\sqrt{2}-1$$
(the $0$'s come from the exponentials with base less that $1$, as they go to zero as the exponent goes to infinity)
No mind what values for $A$ and $B$ we have, so is, the limit is that, irrespective of the values for $x_1$ and $x_2$.
Added
Suppose the sequence has this form $x_n=A·r^n$ and check if it is possible for $r$ and $A$ to meet the conditions the recurrence law imposes:
$A·r^{n+1}=2A·r^n+A·r^{n-1}$ or $A·r^2r^{n-1}=A2·r·r^{n-1}+Ar^{n-1}$
But obviously $A\neq0$ and $r\neq0$, so we can simplify and $r$ must satisfy:
$r^2-2r-1=0$ with roots $1-\sqrt{2}$ and $1+\sqrt{2}$
But the equation is linear, thus a linear combination of solutions is too a solution:
$x_n=A(1-\sqrt{2})^n+B(1+\sqrt{2})^n$
A: We have $x_n$ to be monotonically increasing and since $\dfrac{x_{n+1}}{x_n}=2+\dfrac{x_{n-1}}{x_n}$, we can say $\dfrac{x_{n+1}}{x_n}<3$ as $\dfrac{x_{n-1}}{x_n}<1$ (monotoniclly increasing)$\,\:\forall\:n\geq1$. So $x_n$ is convergent.  
Let $x_n=A\cdot a^n,\:a\neq0$. The recurrence relation becomes $$ A\cdot a^{n+1}=2A\cdot a^n+A\cdot a^{n-1}\implies A\cdot  a^{n-1}(a^2-2a-1)=0\underset{\substack{A\neq0\\a\neq0}}{\implies} a^2-2a-1=0$$
By quadratic formula, $a=\dfrac{2\pm\sqrt{4+4}}{2}=1\pm\sqrt{2}$. Since $x_n$'s are non negative, we have $a=\displaystyle \lim_{n\to\infty}x_n=1+\sqrt{2}$
$\rule{17cm}{0.5pt}$
$x_n=A(1+\sqrt{2})^n+B(1-\sqrt{2})^n$ where $A,B$ are independent of $n$ by linear recurrence. 
So if $L=\displaystyle\lim_{n\to\infty}\dfrac{x_n}{x_{n+1}}=\lim_{n\to\infty}\dfrac{A(1+\sqrt{2})^n+B(1-\sqrt{2})^n}{A(1+\sqrt{2})^{n+1}+B(1-\sqrt{2})^{n+1}}=\lim_{n\to\infty}\dfrac{A+\dfrac{B(1-\sqrt{2})^n}{B(1+\sqrt{2})^n}}{\dfrac{A(1+\sqrt{2})^{n+1}}{(1-\sqrt{2})^n}+\dfrac{B(1-\sqrt{2})^{n+1}}{B(1+\sqrt{2})^n}}=\dfrac{A}{A(1+\sqrt{2})}=\sqrt{2}-1$.
$\left(\text{Since}\: |1-\sqrt{2}|<1\implies \displaystyle\lim_{n\to\infty}(1-\sqrt{2})^n\to0\right)$. 
To show that the limit $\displaystyle\lim_{n\to\infty}\dfrac{x_n}{x_{n+1}}$ exists  we see that it is bounded since $$0<\dfrac{x_n}{x_{n+1}}=\dfrac{x_n}{2x_{n}+x_{n-1}}\leq\dfrac{x_n}{2x_{n}}=\dfrac12 $$
and $x_{n+1}\geq2x_n\implies x_{n+1}\geq x_n$ which gives $\dfrac{x_n}{x_{n+1}}=\dfrac{2x_{n-1}+x_{n-2}}{2x_{n}+x_{n-1}}\leq\dfrac{2x_{n-1}+x_{n-2}}{2x_{n}}$ i.e.$\dfrac{x_n}{x_{n+1}}-\dfrac{x_{n-1}}{x_{n}}\leq\dfrac{x_{n-2}}{2x_{n}}$ which shows monotonicity.Hence the limit exists.
A: One trick is to use the given limit to derive the existence of it.
Write $y_n=\frac{x_n}{x_{n+1}}$. 

Claim: There exists $r\in (0,1)$ such that 
  $$(\sqrt 2-1)-y_n\le r\big(y_{n-1}-(\sqrt2-1)\big).$$

Proof: Rearranging the terms,
\begin{align*}
\underbrace{\frac{x_{n+1}}{x_n}}_{=1/ {y_n}}-1&=\underbrace{\frac{x_{n-1}}{x_n}}_{=y_{n-1}}+1\\
\frac 1{y_n}-(\sqrt2+1)&=y_{n-1}-(\sqrt2-1)\\
\frac 1{y_n}-\frac1{\sqrt2-1}&=y_{n-1}-(\sqrt2-1)\\
(\sqrt 2-1)-y_n&=\underbrace{y_n(\sqrt2-1)}_{\in(0,1)}\big(y_{n-1}-(\sqrt2-1)\big).
\end{align*}
so the claim is true. $\square$
Therefore,
\begin{align*}
|y_n-(\sqrt2-1)|&\le r |y_{n-1}-(\sqrt2-1)|\le r^2 |y_{n-2}-(\sqrt2-1)|\le\cdots\\
&\le r^{n}|y_0-(\sqrt2-1)|\to0,
\end{align*}
implies the limit exists, and $\lim\limits_{n\to\infty}y_n=\sqrt2-1$.
A: Look at $x_n/x_{n+1} - x_{n-1}/x_n$ which is $(x_n^2 - x_{n-1}x_{n+1})/x_n x_{n-1}$. 
We can prove by induction that the numerator is $(-1)^n$.
$$x_{n+1}^2 - x_n x_{n+2}
= x_{n+1}^2 - x_n(2x_{n+1} + x_n)
= x_{n+1}(x_{n+1} - 2x_n) - x_n^2
= -(x_n^2 - x_{n-1}x_{n+1})$$
with $x_1^2 - x_0x_2 = -1$. Hence $x_n/x_{n+1}$ tends to a limit by the alternating series test.
A: Since $x_n$ is increasing, all the terms become non-zero. By defining $a_n={x_n\over x_{n+1}}$ we have $${x_n\over x_{n+1}}={x_n\over 2x_n+x_{n-1}}={1\over 2+{x_{n-1}\over {x_n}}}$$therefore $$a_n={1\over 2+{ a_{n-1}}}$$Now by defining $b_n=a_n-(\sqrt 2-1)$ we have $$b_n+\sqrt 2-1={1\over b_{n-1}+\sqrt 2+1}$$therefore $$b_n=-\sqrt 2+1+{1\over b_{n-1}+\sqrt 2+1}={(1-\sqrt 2)b_{n-1}\over a_n+2}$$since $x_n>0$ we have $a_n>0$ therefore$$|b_n|=|{(1-\sqrt 2)b_{n-1}\over a_n+2}|\le {\sqrt 2-1\over 2}|b_{n-1}|$$which means that $b_n \to 0$ or $a_n={x_n\over x_{n+1}}\to \sqrt 2-1 \quad\blacksquare$
A: 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.
