Monotonic sequence and limit 
For $x_n = \frac{1}{2+x_{n-1}}$ where $x_1 =1/2$, show that the
  sequence is monotonic and find its limit.

What I first did was finding $x_{n+1}$, which equals $\frac{1}{2+x_n}$; then $x_{n+2}=\frac{1}{2+\frac{1}{2+x_n}}=\frac{x_n+2}{2x_n+5}$ thus it does eventually get smaller hence $x_n>x_{n+1}$. How can I finish this? 

We prove by induction that the odd subsequence is decreasing and that the even subsequence is increasing.
Proceed by observing that $x_{1} > x_{3}$ and $x_{2} < x_{4}$. Now assume that $$x_{2n-1} > x_{2n+1}, \quad x_{2n} < x_{2n+2}$$
Now we have
$$ x_{2n+3} = \frac{1}{2 + x_{2n+2}} < \frac{1}{2 + x_{2n}} = x_{2n+1}$$
With this, we have
$$ x_{2n+2} = \frac{1}{2 + x_{2n+1}} < \frac{1}{2 + x_{2n+3}} = x_{2n+4} $$
Thus, by induction we have our claim.
Also, prove by induction that $x_{2n} < x_{1}$ and $x_{2n-1} > x_{2}$ for all $n \in \mathbb{N}$.
Thus both the even and the odd subsequence converge. We need to show that they converge to the same value.
Now assume that the limit exists (justified below) and is equal to $L$. Then
$$L = \lim x_{n} = \lim\frac{1}{2 + x_{n-1}} = \frac{1}{2 + L}$$
from where we get that
$$ L = (\pm\sqrt{2} - 1)$$
But $L > 0$ (Why?) and hence $L = \sqrt{2} - 1$.
EDIT: For the existence of the limit, see the Edit in Did's proof.
