Proof Verification: Show sequence is bounded and find limit: $x_1 \gt 1$ and $x_{n + 1} = 2 - \frac{1}{x_n}$ Came across the following exercise in Bartle's Elements of Real Analysis and am a little unsure about my solution. Would be extremely grateful if someone could verify it for me.  

Let $x_1 \in \Bbb R$ satisfy $x_1 \gt 1$ and let $x_{n + 1} = 2 -
\frac{1}{x_n}$ for $n \in \Bbb N$. Show that the sequence $(x_n)$ is
  monotone and bounded. What is the limit?

My Attempt: 
Part 1 - (Showing monotony and bound)
We will prove that $1 \lt x_{n +1} \lt x_n$ for each $n \in \Bbb N$. $x_2 = 2  - \frac 1 {x_n} \gt 2 - 1 = 1$ and $x_2 = 2  - \frac 1 {x_2} = \dfrac{2x_2 - 1}{x_2} = 1 + \dfrac{x_2 -1}{x_2} \lt 1 + x_2 - 1 = x_2$ whence we have that $1 \lt x_2 \lt x_1$. Now suppose $1 \lt x_{n +1} \lt x_n$ for an arbitrary natural number $n$. 
$$x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } \gt 2 - 1 = 1$$ 
$$ x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } = \frac{2x_{n + 1} - 1}{x_{n + 1}} = 1 + \frac{x_{n + 1} - 1}{x_{n + 1}} \lt 1 + x_{n + 1} - 1 = x_{n + 1} $$
Therefore we have $ 1 \lt x_{n + 2} \lt x_{n + 1} $ and hence induction is complete. 
So the sequence is monotone decreasing and is bounded below by $1$. By the Monotone Convergence Theorem $\lim (x_n) = x$ exists in $\Bbb R$. 
Part 2 - (Finding Limit)
We know that $x_nx_{n + 1} = 2x_n - 1$ and that $(x_n)$ converges and so does $(x_{n + 1})$ to the same limit since it can be interpreted as a subsequence of $(x_n)$. Then the sequence $(x_nx_{n + 1})$ will also converge to the limit $\lim {(x_n)} \lim {(x_{n+ 1})} = x^2 $. The sequences $(y_n)$ and $(z_n)$ given by $y_n = 2$ and $ z_n = -1 $ for $n \in \Bbb N$  also converge to $2$ and $(-1)$ respectively so $\lim (2x_n - 1) = 2x - 1$. Whence we have that $$x^2 = 2x - 1 \implies x = 1$$


  
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*Am a little unsure  about two areas of the proof - one, the part which uses induction since I found both steps can be shown using the
  same method and two, the computation of $\lim (x_n)$. So any criticism
  is more than welcome. 
  
*If there is an easier way to prove this I would very much like to know since this is just the first exercise and I do not want to follow
  a wrong path.
  


To take this out of unanswered list:
We will prove that $1 \lt x_{n +1} \lt x_n$ for each $n \in \Bbb N$. $x_2 = 2  - \frac 1 {x_n} \gt 2 - 1 = 1$ and $x_2 = 2  - \frac 1 {x_2} = \dfrac{2x_2 - 1}{x_2} = 1 + \dfrac{x_2 -1}{x_2} \lt 1 + x_2 - 1 = x_2$ whence we have that $1 \lt x_2 \lt x_1$. Now suppose $1 \lt x_{n +1} \lt x_n$ for an arbitrary natural number $n$. 
$$x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } \gt 2 - 1 = 1$$ 
$$ x_{n + 2} = 2 - \frac{1}{x_{n+ 1} } = \frac{2x_{n + 1} - 1}{x_{n + 1}} = 1 + \frac{x_{n + 1} - 1}{x_{n + 1}} \lt 1 + x_{n + 1} - 1 = x_{n + 1} $$
Therefore we have $ 1 \lt x_{n + 2} \lt x_{n + 1} $ and hence induction is complete. 
So the sequence is monotone decreasing and is bounded below by $1$. By the Monotone Convergence Theorem $\lim (x_n) = x$ exists in $\Bbb R$. 
We know that $x_nx_{n + 1} = 2x_n - 1$ and that $(x_n)$ converges and so does $(x_{n + 1})$ to the same limit since it can be interpreted as a subsequence of $(x_n)$. Then the sequence $(x_nx_{n + 1})$ will also converge to the limit $\lim {(x_n)} \lim {(x_{n+ 1})} = x^2 $. The sequences $(y_n)$ and $(z_n)$ given by $y_n = 2$ and $ z_n = -1 $ for $n \in \Bbb N$  also converge to $2$ and $(-1)$ respectively so $\lim (2x_n - 1) = 2x - 1$. Whence we have that $$x^2 = 2x - 1 \implies x = 1$$
