Why is $x^n \psi^{(m)}(x)$ zero at infinity? 
Prove: Product of any polynomial of $x$ with $\psi(x)$ or any of its derivatives goes to zero in the limit $x\to\pm\infty$.

This comes from a footnote written by the professor in his quantum mechanics problem set and I am curious to know why. I have unsuccessfully attempted L'Hôpital, Taylor expansion, and many other weird methods. Any help is appreciated.  
 A: A counterexample where one can provide an explicit Hamiltonian is the case of a particle under a constant force (linear potential).
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2} \psi(x) = F\cdot \,(x-x_0)\,\psi(x)\,.
$$
The solution is famously given by the Airy function.
$$
\psi(x) = c_A \mathrm{Ai}(\sqrt[3]{F}\cdot(x-x_0)) + c_B \mathrm{Bi}(\sqrt[3]{F}\cdot(x-x_0))\,.
$$
If $x < x_0$ and $|x-x_0|\gg1$, then, letting $z= - \sqrt[3]{F}\cdot(x-x_0)$,
$$
\mathrm{Ai}(-z) \sim  \frac{\sin \left(\frac23z^{\frac{3}{2}}+\frac{\pi}{4} \right)}{\sqrt\pi\,z^{\frac{1}{4}}} \left(1 + O(z^{-3/2})\right)\,.
$$
This clearly does not go to zero when multiplied by a polynomial.
A: In general, not all states in the Hilbert space satisfy your professor's requirement. One guess for what your professor had in mind is that they wanted the wave function to be "nice" enough that the $\hat{X}$ and $\hat{P}$ operators can act on the state arbitrarily many times while still outputting a physically valid state in the Hilbert space.
In general, the $\hat{X}$ and $\hat{P}$ operators are not closed on the Hilbert space: there exist wave functions that are square-integrable, but if you act $\hat{X}$ on them enough times (i.e. multiply by them by a high enough power of $x$), then the result is no longer square-integrable. So technically, the $\hat{X}$ and $\hat{P}$ operators are not actually linear operators on the entire Hilbert space $\mathcal{L}_2(\mathbb{R}^n)$. But they are closed on a (non-complete) subspace of the Hilbert space called the Schwartz space. It's pretty straightforward to show that the Schwartz space respects your professor's claim. Your professor may have been considering this special subset of "nice" wave functions, but a general state doesn't meet the stated requirement.
