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!

Firstly, I'd like to thank everybody who contributed! Your insight helped me greatly in understanding the problem. Here is my solution (one that I feel is more down to earth in terms of mathematical skills used!)
It is easy enough to prove that $x_n\in[0,1] \forall n$ through induction. Now we definitely have a bounded sequence.
However, it is not monotonic and after some investigation, one can see that it seems to oscilate around its limit $\frac{\sqrt{5}-1}{2}$ - that is, if it actually converges!
Now the crucial step. I used the expression $x_{n+2}=\cfrac{1}{1+\cfrac{1}{1+x_n}}$ to check for which values it is monotonic, which lead to this:
$x_{n+2}>x_n$  for $x_n\in(0,\frac{\sqrt{5}-1}{2})$
$x_{n+2}<x_n$ for $x_n>\frac{\sqrt{5}-1}{2}$
Thus we have two subsequences, both of which are bounded and monotonic and converging to the same limit, only from two different sides.
This answer isn't probably full, but the additional proofs needed are trivial enough, this was the most important step I needed to take to get to the answer.
