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.