I tested this limit on WolframAlpha, $$ \lim_{t\to\infty}\frac {2\ 3^r (2 t - 1) - 6} {3\ 2^r (2 t - 1) - 6}=\left(\frac{3}{2}\right)^{r-1},$$
which displayed two contour maps:
.
Can someone describe the what we are seeing? Anything notable?
Edit Question has been answered, but I've added motivaton. In Mathematica, when we take the limit to a specific triangle $t$, we get a generating function for that triangle that produces all numbers that need to be proved to always be fractions, in order to prove there are no loop-backs in Collatz. Say we set $t\rightarrow2$, we get $\frac{2 \left(3^r-1\right)}{3\ 2^r-2}$, then we get $\{\frac{8}{5},\frac{26}{11},\frac{80}{23},\frac{242}{47},\frac{728}{95},\frac{2186}{191},\frac{6560}{383},\frac{1514}{59},\frac{59048}{1535}\}$ for the first few $r$'s. If we can prove this for all triangles, we can show that no looping occurs in Collatz.
This limit is rather simple, being of the form $lim_{t\to \infty}(at+b)/(ct+d)=a/c$. The fact that $r$ and $t$ be integer and $t$ a multiple of $r$ has no impact, there is still an accumulation point at $\infty$.
The plot is of little use, but the 3D view clearly shows the exponential behavior as a function of $r$, just perturbed by the zeroes of the denominator. These appear as an exponential curve on the 2D diagram but are irrelevant for the limit computation.