# Thread: Limit 1

1. ## Limit 1

Sequences $\{x_n\}_{n=1}^\infty$ is defined with
$\displaystyle

Find $\lim\limits_{n\to\infty}x_n$

2. The possible finite limits of the sequence are the solutions of the equation...

$x_{n}=x_{n-1} \rightarrow x^{2} -2\cdot x -2 =0$ (1)

... that are...

$x= 1 \pm \sqrt{3}$ (2)

A quickly inspection reveals that if $x_{0}= \frac{1}{3}$ is...

$\lim_{n \rightarrow \infty} x_{n} = 1- \sqrt{3} = -.732050807568 \dots$ (3)

Kind regards

$\chi$ $\sigma$

3. Yes, that is limit. But how to prove convergence of this, using therem of monoton and bounded sequences? It's not quite easy, it must be considered, I think, subsequences
$x_{2n}$ and $x_{2n-1}$
One is increasing, second decreasing, and both oscilate about $1-\sqrt 3$. This may be a strict proof.

4. It is not at all difficult to prove that...

$x_{n+1} > x_{n}$ if $x_{n} > 1 + \sqrt {3}$ or $x_{n}< 1 - \sqrt{3}$

$x_{n+1} = x_{n}$ if $x_{n} = 1 \pm \sqrt {3}$

$x_{n+1} < x_{n}$ if $1-\sqrt{3} < x_{n} < 1 + \sqrt {3}$ (1)

On the basis of these unequalities we have the following distinct cases...

a) if $x_{0} > 1 + \sqrt{3}$ the sequence diverges at $+ \infty$

b) if $x_{0} = 1 + \sqrt{3}$ is $x_{n}= 1 + \sqrt{3}$ , $\forall n$

c) if $x_{0} < 1 + \sqrt{3}$ the sequence converges at $1 - \sqrt{3}$ in any case with 'oscillatory behavior' unless is $x_{0} = 1-\sqrt{3}$ that produces the constant sequence $x_{n} = 1 - \sqrt{3}$ , $\forall n$ ...

Kind regards

$\chi$ $\sigma$

5. $1-\sqrt 3=-0.73205...$

$x_1=-0.94444...$,
$x_2=-0.55401...$,
$x_3=-0.84653...$,
$x_4=-0.64168...$,
$x_5=-0.79411...$,
$x_6=-0.68468...$,
$x_7=-0.76560...$,
$x_8=-0.70692...$,
$x_9=-0.75012...$,
$x_{10}=-0.71865$
$\dots$

and you can "see " that $x_{2n-1}>1-\sqrt 3$, and "is" monotonicaly increasing, and $x_{2n}<1-\sqrt 3$, and "is" monotonicaly decreasing.
So, here is neccessary only strict proof of this monotonity. Maybe this relation
$
x_{n+1}-x_n=\frac 12\left[x_n-(1-\sqrt 3)\right]\left[x_n-(1+\sqrt 3)\right ]
$

Observe that $-1, (by inicial value) for all $n=1,2,3\dots$, so second parenteze is negative, and somehow here must be used induction.
I belive similar can be done for
$x_{2n+2}-x_{2n}$ and $x_{2n+1}-x_{2n-1}$

Update: Hear is another relation
$x_{n+2}-x_n=\frac {1}{4}\, (x_{n+1}-x_n)[\underbrace{(x_n+1)^2+1}_{> 0}]$, so this could be used maybe (I think) for variant of induction (with assumptation for $n-1$ and $n$, and proff for $n+1$ and $n+2$).