Why do the limit of $\frac{d\Psi}{dx}$, $\Psi\frac{d\Psi}{dx}$, and $ x\Psi\frac{d\Psi}{dx} \to 0$ as $x \to \infty$?

Question

When showing that $\Psi$ stays normalised, we arrive at

$$ \frac{d}{dt} \int_{-\infty}^{+\infty} \lvert \Psi \rvert^2 dx = \frac{i\hbar}{2m} \int_{-\infty}^{+\infty} \frac{\partial}{\partial x} \bigg( \Psi^* \frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x} \Psi \bigg) dx $$

$$ = \frac{i\hbar}{2m} \Bigg[ \Psi^* \frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi \Bigg]_{-\infty}^{+\infty} = 0 $$

We claim that this equals $0$ because

$$ \lim_{x \to \pm\infty} \bigg( \Psi^* \frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi \bigg) = 0 $$

since $\displaystyle \lim_{x \to \pm\infty} \Psi = 0$ and $ \displaystyle \lim_{x \to \pm\infty} \frac{\partial\Psi}{\partial x} = 0 $, and these because $\displaystyle \int_{-\infty}^{+\infty} \lvert \Psi \rvert^2 dx = 1$ is finite. $\ $ (Griffiths, 1.4)

This may be pedantic, but it’s not entirely clear to me...

Question 1.

Why the $ \lim_{x \to \infty} \Psi$ or $ \frac{d\Psi}{dx} \to 0$.

I can come up with $\Psi$ such that the above ($\Psi \to 0 $) is not technically true in the sense that it is Not true that $\forall \epsilon > 0\ \exists x_0 $ such that $ x > x_0 \implies \lvert \Psi(x) - 0 \rvert < \epsilon $.

Consider, for example, a sequence $\{ a_i \}$ whose sum converges to $1$. Then construct a $\Psi(x)$ which is $0$ everywhere, except for spikes, centred at integers $i$, each of area $a_i$.

Furthermore, if the spikes are made narrower as their area decreases, but not shorter, then $\frac{d\Psi}{dx}$ doesn’t approach $0$ either, but rather, in fact, a monotonically increasing series can be found.

Question 2.

Even if $\displaystyle \int_{-\infty}^{+\infty} \lvert \Psi \rvert^2 = 1 \implies \lim_{x \to \pm\infty} \Psi = 0 $ or $\displaystyle \lim_{x \to \pm\infty} \frac{\partial\Psi}{\partial x} = 0$, it seems circular to reason that since the integral is finite (ie, $= 1$), the integral must be constant.

Question 3.

When deriving the momentum, we get $$ p = \frac{d \langle x \rangle}{dt} = \frac{i\hbar}{2m} \int_{-\infty}^{+\infty} x \frac{\partial}{\partial x} \bigg( \Psi^* \frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x} \Psi \bigg) dx $$

Evaluating the term

$$ \Bigg[ x\Psi^* \frac{\partial\Psi}{\partial x} - x\frac{\partial\Psi^*}{\partial x}\Psi \Bigg]_{-\infty}^{+\infty} \ \text{as} = 0 $$

Which Griffiths section 1.5 claims "on the ground that $\Psi$ goes to zero at ($\pm$) infinity. " Assuming that both $\Psi \to 0$ and $\frac{\partial\Psi}{\partial x} \to 0$ as $x \to \pm\infty$, it's not clear to me why $\displaystyle \lim_{x \to \infty} x\Psi = 0$ as this is "like" $\infty \times 0$. I tried using L'Hôpital's rule, but didn't get anywhere. And it's not obvious to me why it should.

So why are the limits of these three terms, $\displaystyle \Psi, \frac{\partial \Psi}{\partial x}$, and $\displaystyle x \Psi \frac{\partial \Psi}{\partial x}$ equal to $0$ as $x \to \infty$? And how do we prove that the integral is finite using that the integral is finite? (Why is it not circular)? Perhaps we could instead argue that the two terms in $ \Psi^* \frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi$ cancel if they are both real?