If $x_0=1$ and $x_n=\frac {1}{1+x_{(n-1)}}$, find: $\lim_{x\to\infty} x_n$ 
If $x_0=1$ and $x_n=\dfrac {1}{1+x_{(n-1)}}$, find $\displaystyle\lim_{x\to\infty} x_n$.

My attempt:
$x_1=1+\dfrac 1 2=\dfrac 3 2$
$x_2=1+\dfrac {1}{1+\frac 3 2}=\dfrac2 5$
Which gives following series:
$$1, \frac32, \frac35,\frac 58, \frac 8 {13}, \dots$$
Since the denominator is greater than the numerator, the limit of $x_n=0$ when $x\to\infty$. I do not think this result is correct. I think this problem must have a solution using continued fractions. Any idea?
 A: You miscalculated the two first initial terms of the sequence.
$x_{1}=\frac{1}{1+x_{0}}=\frac{1}{1+1}=\frac{1}{2}$ and $x_{2}=\frac{1}{1+x_{1}}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}$.
Let's try to apply the Fixed Point Theorem. Consider $I=[\frac{1}{2},1]$ and $f(x)=\frac{1}{1+x}$ defined in $I$.

*

*$f$ is continuous in $I$

*$f(I)\subset I$ once $f'<0$

*$f$ is Lipschitz continuous because as $f$ is a function of class $C^1$ you have that $|f'(x)|=\frac{1}{(1+x)^2}\leq \frac{1}{1+\frac{1}{2}}=\frac{2}{3}$. Then for any $x,y \in I$ $\Rightarrow$ $|f(x)-f(y)|\leq\frac{2}{3}|x-y|$.

Then $x_{n}$ converges for the only fixed point of $f$ in $I$, i.e., $l=\frac{1}{1+l}$.
A: First, try to show that a limit even exists in the first place. Then, in doing so, let
$$L := \lim_{n \to \infty} x_n$$
With this, then, taking the limit of both sides of your equation gives us
$$L = \frac{1}{1+L}$$
This will give you two possible values for $L$. Use $x_0$ and a convergence argument to determine which $L$ is the value relevant for your case.

You are right in that this references continued fractions in a sense. If you imagine expanding the recursion backwards, you'll get
$$L = \cfrac{1}{1+ \cfrac{1}{1+ \cfrac{1}{1+ \cfrac{1}{1+ \ddots}}}}$$
There is a self-similarity here: namely everything below the second numerator is in fact just $L$ itself:
$$L = \frac{1}{1+L}$$
However, this is not a fully rigorous way to handle the problem. For instance, it tells you possible values of $L$, but not which ones actually are valid - or, indeed, if any of them are. (Heck, the value of $x_0$ is important, even: there's a value for $x_0$ in this case in which you would get a different $L$ than in your case and in nearly every other case.) An infinite process like this is almost always defined in terms of a limit of the smaller, finite process. For instance, like how we define
$$\sum_{k=1}^\infty a_n := \lim_{n \to \infty} \sum_{n=1}^k a_n$$
We'd first have to know that the limit in question even exists in the first place before assigning the infinite sum a value via any other means. Such is mathematical rigor.
I give a further elaboration on this sort of notion in this answer which ties into continued fractions (particularly the well-known one for $\varphi = (1 + \sqrt 5)/2$ and its conjugate), how we define these, and the convergence of a continued fraction's "partial fractions."
A: It is very simple Fibonacci squence
$\lim_{n\to+\infty} x_n=\lim_{n\to+\infty}\frac{F_{n-1}}{F_n}=\frac{a}{a}=1$
$a^2+a=1$;
$ F_1=1$ ;
$F_{n+1}=F_n+F_{n-1}$
