Treat fixed points as candidate limits before proving decrease

If the sequence converges to a finite limit $s$, the recurrence gives $s=12/(7-s)$, hence $s^2-7s+12=(s-3)(s-4)=0$ and $s=3$ or $4$. A limit $s=7$ is impossible because the next term would become unbounded. Without an initial value or a convergence argument, these are candidate limits, not a proof that every iteration converges.

Let $h(x)=12/(7-x)-x$. The source sketch has zeros at $3,4$, a vertical asymptote $x=7$, and a shallow negative minimum near $(3.54,-0.0718)$. The exact minimum on the left branch is $(7-2\sqrt3,4\sqrt3-7)$. The graph is below zero for $3<x<4$, so $h(x_n)=x_{n+1}-x_n<0$.

$h(x)=(x^2-7x+12)/(7-x)=(x-3)(x-4)/(7-x)$. In $(3,4)$ the numerator is negative and denominator positive, proving strict decrease. Also $3<12/(7-x)<4$ there. Thus if $3<x_1<4$, the terms remain in that interval, decrease and are bounded below by $3$; their actual limit is $3$. The constant sequence $x_n=4$ provides the other fixed-point limit.