How is classical mechanics recovered when the commutator is zero? If $X$ and $P$ commute, then the rate of change of expectation value of $X$ becomes zero,  assuming
$$\frac{d}{dt} \langle X \rangle= \langle [X, P^2+V(x)] \rangle=0.$$
This is not what classical mechanics says, is it?
 A: One has to be careful in discussing the transition from quantum to classical mechanics.  First, by Dirac quantization (see also this post):
$$
[\hat A,\hat B]\to i\hbar \{A,B\}_{PB} +{\cal O}(\hbar^2) \tag{1}
$$
where $\{A,B\}_{PB}$ is the Poisson bracket.  Thus, if you naively set $\hbar\to 0$, you get nonsense.  In particular you have no dynamics as this comes out of the Poisson bracket of a function and the Hamiltonian.  Note that, in (1), the left hand side refers to the commutator of operators whereas the right hand side refers to the PB of functions in phase space (of $p$ and $q$).
Within the formalism of Wigner quasidistributions, which is probably the most natural to investigate the quantum-classical transition, the classical limit is not obtained by setting $\hbar=0$ but by ignoring higher powers of $\hbar$ past the Poisson bracket in the expansion of the Moyal bracket.
Even in the WKB formalism (which is an expansion in $\hbar$), the leading term, from which we extract the lowest order WKB approximation, still contains one power of $\hbar$.
Thus recovering classical mechanics from quantum mechanics is a subtle business it is misleading to suggest that the classical mechanics is obtained by simply setting $\hbar\to 0$.
A: Assuming $X$ is the position operator, we cannot interpret $P$ as a canonical momentum operator conjugate to $X$ because they commute. Therefore, the quantity $P^2 + V(x)$, which I assume is your Hamiltonian, has no kinetic term for the $X$ direction, so no dynamics in $X$.
This argument applies to classical mechanics too if you use the Poisson bracket instead. So if we let $\{x,p\} = 0$, then
$$ \frac{dx}{dt}= \{x,H\} = \{x,p^2 + v(x)\} = 0 $$
A: I got it
The formula $i\frac{d}{dt} \langle X\rangle =\langle [X,H]\rangle $ is arrived at using the schrodinger equation, and hence is only valid in quantum mechanics. The equivalent formula for classical mechanics would be derived using the matrixified version of Hamilton's equations.
