Consider Ehrenfest's theorem: \begin{align} m\frac{d\langle x\rangle}{dt}=\langle p\rangle \\ \frac{d\langle p\rangle}{dt}=-\langle V'(x)\rangle. \end{align} Suppose $V(x)=x^2+x^{n+1}$ where $n>1$. Then $\langle V'(x)\rangle\neq V'(\langle x \rangle)$. Therefore the evolution of $\langle x\rangle$ and $\langle p\rangle$ is different to the classical prediction: recall that classically \begin{align} m\dot{x} &= p \\ \dot{p} &=- V'(x). \end{align} Now, we are told that if $\hbar\to 0 $ quantum mechanics reproduces classical physics. Therefore I would expect that $\langle x^n \rangle-\langle x \rangle^n$ is $O(\hbar)$ (because it is this term which is responsible for the deviation of the evolution of $\langle x\rangle$ and $\langle p\rangle$ from the classical prediction). However, from the quantum mechanical formulation, I don't see why this term is actually $O(\hbar)$.
So my question is: is $\langle x^n \rangle-\langle x \rangle^n$ actually $O(\hbar)$? If it isn't then how do I explain why $\hbar\to 0$ reproduces classical predictions for the evolution of $\langle x\rangle$ and $\langle p\rangle$?
[Notes: For one, the smallness of $\langle x^n \rangle-\langle x \rangle^n$ depends on the details of the state: if $\psi(x)=\delta(x-x_0)$ for example then $\langle x^n \rangle-\langle x \rangle^n=0$. But the classical limit should not depend on the details of the state, there must be some general arguments for why $\langle x^n \rangle-\langle x \rangle^n$ is small -- my own feeling is that it is controlled by $\hbar$, but, as my question says, I can't see why.]
