How does the delta prime boundary conditions behave when the jump factor approaches infinity?

I recently came across the concept of a $delta'(x)$ (delta prime) potential, which is basically a potential which imposes the boundary condition:

• $frac{partialpsi}{partial x}$ is ‘continuous’ at $0$, in the sense that $frac{partialpsi}{partial x}|_{0^+}=frac{partialpsi}{partial x}|_{0^-}=$:$psi'(0)$, meaning that both one sided limits exist and are equal (although $psi$ doesn’t have to be differentiable in the usual sense).
• The double sided limits of $psi (0)$ exist and we have –
  $$psi(0^+)-psi(0^-)=psi'(0)$$

Note that $psi$ doesn’t have to be continuous at $0$.

I wish to look at the potential of the form $-sigmadelta'(x)$ (where $sigmainmathbb R$), which is defined just as before, only this time we have the condition:
$$psi(0^+)-psi(0^-)=-sigmapsi'(0)$$

This gives me a certain boundary condition for any choice of $sigma$. My question is – what is the condition as $sigmarightarrow infty$? In a sense – what happens if I take $sigma=infty$? Is there a ‘proper’ way to describe this boundary condition?

It seems to me that for me to take this limit, I must have in the corresponding boundary condition that $psi'(0)=0$. But what I’m interested in is – does $psi$ now have to be continuous at $0$? We now have something of the form:
$$psi(0^+)-psi(0^-)=-inftycdot 0$$
Naively, we can’t estimate the RHS and claim if $psi$ is continuous at $0$. But maybe anyone here knows of a ‘proper’ way to take this limit (via distribution theory or something) so that we can know if maybe $psi$ also needs to be continuous at $0$ for some reason?

I’d be happy to hear all kinds of answers – intuitions, formal proofs (this is probably best), and even references to papers/books which do something similar to what I described.

Thanks in advance!