I am confused, so please help me out.
Background
Consider the dynamical system for $0 \leq x \leq 1$ given by the function
$$f(x) = \begin{cases} \frac 12 (3x+1), & x \in [0, 1/3) \\ \frac 14 (3x-1), & x \in [1/3, 1] \end{cases}$$
Lets call the pieces of $f$ by the names $\sigma(x) = (3x+1)/2$ and $\tau(x) = (3x-1)/4$. Now, by happy coincidence, $\sigma \tau = \tau \sigma$ (function composition), and, by solving simple recurrence relations, we have
$$\sigma^m \tau^n (x) = (1 + x)(3/2)^m(3/4)^n - 1$$
(Note that a simple change of variables elucidates why $\sigma$ and $\tau$ commute: they are just multiplication by a constant!)
OK, so, I want to know how often we use a $\sigma$ on average when we calculate $f^n(x)$ as $n \to \infty$.
Now, I look at the intervals $[0, 1/2]$ and $(1/2, 1]$. We know that $f^n(x) \in (1/2, 1]$ if and only if $f^{n-1}(x) \in [0, 1/3)$, i.e., if and only if the last operation was a $\sigma$ instead of a $\tau$. If we draw $x$ from a uniform distribution on $[0, 1]$, we have that $p_n$, the probability that $f^n(x) \in (1/2, 1]$ is given by the recurrence relation $p_n = 2/3(1 - p_{n-1})$, $p_0 = 1/2$.
This recurrence relation is solved by $p_n = (-2/3)^n/10 + 2/5$, and we clearly have $\lim_{n \to \infty} p_n = 2/5$. Fine so far, right?
Now is the part where I get confused. So, if asymptotically $2/5$ of the operations are $\sigma$, and the other $3/5$ of the operations are $\tau$, then I should expect something like the following for big $n$:
$f^n(x) \approx (1 + x)(3/2)^{2n/5}(3/4)^{3n/5} - 1$
(for almost every $x$).
But, $\lim_{n \to \infty} (3/2)^{2n/5}(3/4)^{3n/5} = 0$, which doesn't make any sense because $f^n(x) \in [0, 1]$, and this is saying that $f^n(x) \to -1$ almost surely.
Question
What is my misunderstanding here?
Motivation
For $x$ a positive integer, let $n$ be such that $2^n \leq x < 2^{n+1}$. Then, we associate with $x$ the dyadic rational $\varphi(x) = (x - 2^n)/2^n$. For odd $x$, this transformation sends $x$ and $2^k x$ to the same point, for any $k \in \mathbb{N}$.
Now, consider the transformation $T(x) = (3x + 1)/2^n, x \equiv a_n \pmod{2^{n+1}}$, where $a_n = 2^n + ((-2)^{n+1}-1)/3$. This is the odd-only version of the transformation from the $3x+1$ problem.
Restricting our attention to odd $x$, for $0 \leq \varphi(x) < 1/3$, we have
$\varphi T(x) = \dfrac{3 \left( \dfrac{x - 2^n}{2^n} \right) + 1}{2} + \dfrac{1}{2^{n+1}} = \dfrac{3 \varphi(x) + 1}{2} + \dfrac{1}{2^{n+1}}$
and for $1/3 \leq \varphi(x) < 1$, we have
$\varphi T(x) = \dfrac{3 \left( \dfrac{x - 2^n}{2^n} \right) - 1}{4} + \dfrac{1}{2^{n+2}} = \dfrac{3 \varphi(x) - 1}{4} + \dfrac{1}{2^{n+2}}$
Thus, if $x, Tx, T^2x, \ldots$ is a divergent trajectory in the $3x+1$ problem, then $n \to \infty$, and the behavior of $\varphi x, \varphi Tx, \varphi T^2x, \ldots$ approaches that of the dynamical system in this question.