Convergence of a sequence to a value and not another. I was tinkering with the following equation and produced an infinite nested fraction:
$$
(x-6)(x-3)=0
$$
$$
x^2-9x+18=0
$$
$$
x=9-\frac{18}{x}
$$
$$
x=9-\frac{18}{9-\frac{18}{9-\frac{18}{...}}}
$$
Clearly:
$$
x=3 \text{ or } x=6
$$
But when we compute the fraction to a finite nuber of terms we notice that:
$$ 
9-\frac{18}{9-\frac{18}{9-\frac{18}{...}}}\to6
$$
Why doesn't it tend to $3$?
Note: I've noticed similar behavior in other fractions and they seem to tend to the greatest of both possibilities. Some thoughts on this will be highly appreciated.

You’ve found that the mapping $x\mapsto f(x)=9-18/x$ has the two fixed points $3$ and $6$. At $3$, the derivative ($18/x^2$) has value $18/9=2$, so that (for small $\varepsilon$), $3+\varepsilon\mapsto f(3+\varepsilon)\sim f(3)+2\epsilon=3+2\epsilon$, that is gets farther from $3$ than it started out. On the other hand, using the same technique, $f(6+\varepsilon)\sim f(6)+\frac12\varepsilon=6+\frac12\varepsilon$, in other words gets closer to $6$. So $6$ is an attractive fixed point, and $3$ is repulsive.
