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?
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 $f'&lt;0$
$f$ 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., $l=\frac{1}{1+l}$.
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$ ;
$F_{n+1}=F_n+F_{n-1}$