I have just started reading through quantum mechanincs (Griffiths), where he says that the below boundary term is zero, as $\Psi$ must go to zero at infinity (when calculating $d \langle x \rangle / dt$) to simplify and do the integration by parts:
$$x\Psi^* \frac{\partial \Psi}{\partial x} - x\frac{\partial \Psi^*}{\partial x}\Psi \ \Bigg|_{-\infty}^{+\infty} = 0$$
At infinity, well, $x$ also goes to infinity making the limit indeterminate. I don't really understand this claim. Can anyone tell me what I'm overlooking here?
Edit1:
So, to be more precise on where my confusion is, we start with
$$\frac{d}{dt} \int_{-\infty}^{+\infty} x(\Psi^*\Psi) dx = \frac{i \hbar}{2m} \int_{-\infty}^{+\infty} x \frac{\partial}{\partial x}\big( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \big)dx$$ which is the result of subbing $\partial\Psi$ for what it is in Schrödinger's equation. From this, we expand it (using integration by parts), resulting in
$$\frac{d \langle x \rangle}{dt} = -\frac{i \hbar}{2m} \int_{-\infty}^{+\infty} \frac{\partial}{\partial x}\big( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \big)dx + \Big(x\Psi^*\frac{\partial \Psi}{\partial x} - x \frac{\partial \Psi^*}{\partial x}\Psi\Big)_{-\infty}^{+\infty}$$
and this is where they argue that the boundary term is zero. I can't see why because that term also has $x$, making the limit indeterminate $0 \times \infty$.