Your teacher has presumably taught you the way out of all these
pseudoparadoxes: Consider Gaussian states of vanishing width. Instead of doing that, effectively hijacking your question to something more interesting, I will remind you of the erroneous way you are multiplying 0 to infinities of different orders!
The short answer is that, technically (physics aside: there is little physics in $|x\rangle$ unnormalizable states!), $\Delta x=0$, indeed, but $\Delta p=\infty$, to satisfy Robertson's formal inequality. You are effectively pleading for a formal way to bypass your technical mistake ("you get"; no you don't!). Take $\hbar=1$ w.l.o.g.
For starters, remind yourself of the identity
$$
\epsilon \delta(\epsilon)=0, ~~~~\leadsto \\
\epsilon \partial_\epsilon \delta(\epsilon)= -\delta(\epsilon).
$$
You are effectively considering matrix elements implicitly under the final step operation $\lim_{\epsilon \to 0}$. Specifically, evaluating the canonical commutator,
$$
\langle x+\epsilon|[\hat x, \hat p]|x\rangle = i\delta (\epsilon),
$$
divergent.
The same follows from your formal pathway,
$$
\langle x+\epsilon|[\hat x, \hat p]|x\rangle =(x+\epsilon)\langle x+\epsilon| \hat p |x\rangle - x \langle x+\epsilon| \hat p |x\rangle=\epsilon \langle x+\epsilon| \hat p |x\rangle\\
=-i\epsilon \int \!\! dy~~ \langle x+\epsilon|y\rangle \partial _y \langle y|x\rangle= -i\epsilon \partial_\epsilon \delta (\epsilon)= i\delta(\epsilon),
$$
by dint of $\hat p=-i \int \!\! dy~~ |y\rangle \partial _y \langle y|$.
So, taking the limit, you indeed find a divergent
$$
\Delta x \Delta p \geq \delta (0)/2~ .
$$
(You can take a further derivative in the leading identity,
$\partial_\epsilon^2\delta (\epsilon) = {2\over \epsilon^2} \delta (\epsilon)$.)
