Find the limiting value of this sequence

Question

Let $x_1=-\sqrt3$ and $x_n=-\sqrt{3-x_{n-1}} $ Evaluate $lim_{n \to \infty}x_n $. How do I do these kind of problems? A kick-off would be highly appreciated.

Answer 1

Hint:

let $\lim_{n\rightarrow \infty}x_n=l$

Then from your equation $$ l=-\sqrt{3-l} $$

hence $l^2=3-l$

solving we get

$$ l=\frac{-1\pm \sqrt{13}}{2} $$

I have a feeling, you should go for $l=\frac{-1- \sqrt{13}}{2} $

as every term of sequence is negative.

Where did I use the initial condition?

I dont know, if it is even required.

Answer 2

First, show the limit exists. What follows is the highlights of this, provide details:

$$\begin{align}&\forall\,n\in\Bbb N\;,\;\;x_n\le 3\iff3-x_n\ge 0\implies\text{inductively, the sequence's well defined}\\{}\\ &\text{Thus}\;\;x_n\le 0\;,\;\;\forall\,n\in\Bbb N\;,\;\;\text{but also}\;\;x_n=-\sqrt{3-x_n}\ge-\sqrt6\\{}\\ &x_1:=-\sqrt 3\ge-\sqrt{3+\sqrt{3}}=-\sqrt{3-x_1}=:x_2\\{}\\ &x_n:=-\sqrt{3-x_{n-1}}\stackrel{\text{Ind.}}\ge-\sqrt{3-x_n}=:x_{n+1}\end{align}$$

From the above, $\;\{x_n\}\;$ converges finitely (why?), say to $\;L\;$ ,so by arithmetic of limits:

$$L\xleftarrow[]{}x_n=-\sqrt{3-x_{n-1}}\xrightarrow[]{}-\sqrt{3-L}\implies L^2+L-3=0\implies$$

$$L_{1,2}=\frac{-1\pm\sqrt{13}}2$$

If you followed so far the above, there's only one possible logical choice for $\;L\;$ .