One slopiness that many teachers are guilty of is teaching this misleading thing:
The position operator $\hat{x}$ has eigenvectors $|x_0\rangle$ that obey
$$
\hat{x} |x_0\rangle = x_0|x_0\rangle
$$
and are represented by distributions on domain of $x$: $\delta(x-x_0)$ for different $x_0$. (WRONG)
The incorrect predictions come when student uses this "representing function" as a simple initial condition to find out how a localized psi function spreads out in time, or to calculate expected average of position.
Let me demonstrate the latter case: calculating expected average of position in such state $|x_0\rangle$ using the standard algorithm, we get
$$
\langle x \rangle = \langle x_0|\hat{x}|x_0\rangle = x_0 \langle x_0|x_0\rangle
$$
It is tempting to put $\langle x_0|x_0\rangle = 1$ now, but this is not correct, because we already said that $|x_0\rangle$ is represented by delta distribution. The expression is just not defined, as the integral
$$
\int \delta(x-x_0)\delta(x-x_0)dx
$$
is not defined (or, sometimes said to be infinite). So, here the slopiness of assuming position operator has eigenvectors leads us to incorrect prediction that there is no expected average of position. Such result would be correct for, say, the Cauchy distribution, but it is incorrect for a localized one we implicitly assume to describe here. For any well-localized psi function around $x_0$, the correct answer is close to $x_0$.
The correct way to handle this is to teach that position operator has no eigenfunctions, but we can assign it improper eigenvectors $|x_0\rangle$ that are however no realizable psi functions. So that fact the very position operator used to define such kets has no expected average for such kets is no problem, because physical kets can never be equal to such kets.