Existence of Limit for $x_{n+1} = 1+1/x_{n}$ I'm attempting to prove that if one defines a sequence with $x_{1} = 1$ and 
$$x_{n+1} = 1+\frac{1}{x_{n}}$$
for all $n >1$ then the limit of $\{x_{n}\}_{n \ge 1}$ exists. 
My Attempt:
Since $\{x_{n}\}_{n \ge 1}$ was defined recursively I immediately thought to try and prove that the sequence is monotone and bounded. The problem is that $\{x_{n}\}_{n \ge 1}$ is an alternating sequence. Its first few values are $1,2,3/2,5/8,\ldots$. I checked boundedness next. Since $x_{1} = 1 \le \frac{1+\sqrt{5}}{2}$ we induct on $n$ and assume $x_{n} \le \frac{1+\sqrt{5}}{2}$. For the $x_{n+1}$ case we find that $x_{n+1} \ge \frac{1+\sqrt{5}}{2}$ which I thought was a little odd... Is there another way I should be attacking this proof or can I fiddle with the sequence a bit to use MCT?

This is a trick I have learned here a few years ago.
Let's say you have a sequence $x_n$ defined by some sort of iteration.
If the limit $\lambda$ of the sequence satisfies a quadratic equation with roots $\alpha, \beta$ and one construct an auxiliary sequence by
$$y_n = \frac{x_n - \alpha}{x_n - \beta}$$
Sometimes, the recurrence relation for $y_n$ will be so simple to the point we can deduce the convergence of $x_n$ or even an closed form expression of $x_n$.
For the problem at hand, the sequence is generated by the iteration.
$$x_{n+1} = 1 + \frac{1}{x_n}$$ 
The limit $\lambda$ need to satisfies a quadratic equation
$$\lambda = 1 + \frac{1}{\lambda}\quad\iff\quad \lambda^2 - \lambda - 1 = 0$$
which have roots $\alpha = \varphi, \beta = -\frac{1}{\varphi}$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.
If one define $\displaystyle\;y_n = \frac{x_n - \varphi}{1 + \varphi x_n}\;$, we have:
$$\begin{align}y_{n+1} 
&= \frac{x_{n+1} - \varphi}{1 + \varphi x_{n+1}}
= \frac{1 + \frac{1}{x_n} - \varphi}{1 + \varphi( 1 + \frac{1}{x_n} ) }
= \frac{(1-\varphi)x_n + 1}{(1+\varphi)x_n + \varphi}\\
&= \frac{1}{\varphi}\frac{(\varphi - \varphi^2)x_n + \varphi}{\varphi^2 x_n + \varphi}
= -\frac{1}{\varphi^2}\frac{x_n - \varphi}{1 + \varphi x_n}
= -\frac{1}{\varphi^2} y_n
\end{align}
$$
Since this is true for all $n \ge 1$, we can solve the recurrence relation and obtain a closed form expression for $y_n$:
$$y_n = \left(-\frac{1}{\varphi^2}\right)^{n-1} y_1 = 
\left(-\frac{1}{\varphi^2}\right)^{n-1} \left(\frac{1-\varphi}{1+\varphi}\right)$$
Since $\left| -\frac{1}{\varphi^2} \right| < 1$, we have $y_n \to 0$ as $n \to \infty$. As a result,
$$\lim_{n\to\infty} x_n = \lim_{n\to\infty} \frac{y_n + \varphi}{1 - \varphi y_n} = \frac{0 + \varphi}{1 - \varphi \cdot 0} = \varphi$$
