<s>[INST] Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?

Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}

 [/INST] Suppose
$$
x = 2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {\ddots}}}}}}
$$
Then
$$
x = 2 + \cfrac 1 {3 + \cfrac 1 x}.
$$
Solving this by the usual method, one gets
$$
x = \frac {3\pm\sqrt{15}} 3.
$$
At this point I'd be inclined to say we obviously want the positive solution. And that that's what the fraction converges to.  However, the answer from Doug M. suggests this point of view: The function
$$
x \mapsto 2 + \cfrac 1 {3 + \cfrac 1 x}
$$
has one attractive fixed point and one repulsive fixed point.
</s>