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? 


The single most useful thing to do here is to draw on a same picture the graphs of the functions $u:x\mapsto1/(2+x)$ and $v:x\mapsto x$, say for $x$ in $(0,1)$. The rest follows by inspection...

Since $u$ is decreasing from $u(0)\gt0$ to $u(\infty)=0$, $u$ has a unique fixed point, say $x^*$. Since $u(x_1)\lt x_1$, one knows that $x_1\gt x^*$. Drawing on our picture the segments from $(x_1,0)$ to $(x_1,x_2)$ to $(x_2,x_2)$ to $(x_2,x_3)$ and so on, one sees that:


*

*the sequence $(x_{2n-1})$ is decreasing and $x_{2n-1}\gt x^*$ for every $n$,

*the sequence $(x_{2n})$ is increasing and $x_{2n}\lt x^*$ for every $n$,

*the whole sequence $(x_{n})$ is neither decreasing nor increasing,

*and the whole sequence $(x_{n})$ converges to $x^*$.


Numerically, $x^*=\sqrt2-1\approx.414$.
Edit: To show the last item, call $L$ the limit of $(x_{2n})$ and $M$ the limit of $(x_{2n-1})$, then $M=u(L)$ and $L=u(M)$ hence $L=u\circ u(L)$ and $M=u\circ u(M)$. Computing $u\circ u$, one sees that $x=u\circ u(x)$ is equivalent to $x=(2+x)/(5+2x)$, that is, $x^2+2x=1$, that is, $x=\pm\sqrt2+1$. Since $L$ and $M$ are positive, this shows that $L=M=x^*$ hence $(x_n)$ converges.
