Finding the limit of $x_1=0 , x_{n+1}=\frac{1}{1+x_n}$ I have had big problems finding the limit of the sequence $x_1=0 , x_{n+1}=\frac{1}{1+x_n}$. So far I've only succeeded in proving that for $n\geq2$: $x_n>0\Rightarrow x_{n+1}>0$
(Hopefully that much is correct: It is true for $n=2$, and for $x_{n+1}>0$ exactly when $\frac{1}{1+x_n}>0$, which leads to the inequality $x_n>-1$ which is true by the induction assumption that $x_n>0$.)
On everything else I failed to come up with answers that make sense (such as proving that $x_{n+1}>x_n \forall n\geq1$). I'm new to recursive sequences, so it all seems like a world behind the mirror right now. I'd appreciate any help, thanks!

Here is a way to show that the sequence converges to the unique positive solution to $a=1/(1+a)$: Define $f(x)=1/(1+x)$. Then $f'(x)=-1/(1+x)^2$. For every $n$, the mean value theorem gives
$$x_{n+1}-a=f(x_n)-f(a)=(x_n-a)f'(c_n)$$
for some $c_n$ between $x_n$ and $a$. Since $-1<f'(x)<0$ for all $x>0$, this shows that $\lvert x_n-a\rvert$ decreases with $n$. Moreover, we already have $f'(1/2)=-4/9$, and $f'$ is increasing, so $-4/9<f'(c_n)<0$ for all $n\ge2$. Thus $\lvert x_n-a\rvert$ decreases at least as fast as $(4/9)^n$, i.e., geometrically. In particular, the $x_n-a\to0$.
Addendum: To do this without differentiation, just rely on the computation
$$f(x)-f(y)=\frac1{1+x}-\frac1{1+y}=\frac{y-x}{(1+x)(1+y)}$$ instead.
