Show that the sequence $(x_k)_k$ is monotone and bounded and find its limit. Let $x_1$ be in $R$ with $ x_1>1$, and let $x_{k+1}=2- \frac{1}{x_k}$ for all $k$ in $N$. Show that the sequence $(x_k)_k$ is monotone and bounded and find its limit.
I am not sure how to start this problem.
 A: by induction $x_k>1$
$x_1 > 1$
If $x_k > 1$ then $\frac {1}{x_{k}} < 1$ and $x_{k+1}  = 2 - \frac {1}{x_k} > 1$
$\{x_k\}$ is montonic
We will show that $x_{k+1} - x_{k} < 0$
$x_{k+1} - x_{k}\\
2 - \frac {1}{x_k} - x_{k}\\
\frac {2x_k -1 + x_k^2}{x_k}\\
-\frac {(x_k - 1)^2}{x_k}$
and since $x_k > 1, -\frac{x_k - 1)^2}{x_k} < 0$
The sequence is monotonic and bounded, therefore convergent.
$\{x_k\}$ converges to $x$
$x = 2 - \frac 1x\\
(x-1)^2 = 0\\
x = 1$
A: Use induction to show that $x_k > 1$ for every $k$. This shows that the sequence is bounded from below.
Next show that the sequence is decreasing by applying induction to prove $x_{k+1}\le x_k$ for every $k$.
Then the monotone convergence theorem implies that the sequence has a limit, say $x:=\lim x_k$. Taking the limit in the relation
$$
x_{k+1}=2-\frac{1}{x_k},
$$
we obtain
$$
x=2-\frac{1}{x}.
$$
Solve this equation for $x$ to find the limit.
A: $$\dfrac12\left(x_{k+1}+\dfrac1{x_k}\right)=1.$$ This says the average of  $x_{2}$ and $\dfrac1{x_1}$ is equals to one. Now suppose $1\lt x_1.$ By the AM-GM inequality $x_1+\dfrac1x_1\gt2$ and this implies $$\text{distance}(x_1,1)\gt\text{distance}\left(1,\dfrac1x_1\right)=\text{distance}(x_2,1).$$ Hence $1\lt x_2\lt x_1.$ Same reasoning shows that $(x_k)_{k\in\Bbb{N}}$ is monotone decreasing.   
