What accounts for a Lyman-break for all wavelengths shorter than 91.2nm if the Lyman limit is the highest energy photon that Neutral hydrogen absorbs?

Question

From this description of Lyman-break galaxies, I don't understand how:

...radiation at higher energies than the Lyman limit at 912 Å is almost completely absorbed by neutral gas around star-forming regions of galaxies. In the rest frame of the emitting galaxy, the emitted spectrum is bright at wavelengths longer than 912 Å, but very dim or imperceptible at shorter wavelengths—this is known as a "dropout", or "break".

But the wikipedia page for the Lyman series states that :

the Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n ≥ 2 to n = 1

and:

The greater the difference in the principal quantum numbers, the higher the energy of the electromagnetic emission.

It then states that:

There are infinitely many spectral lines, but they become very dense as they approach n = ∞ (the Lyman limit)... "91.1753 nm"

I don't understand how neutral hydrogen can absorb light from "a photon emitted with wavelength shorter than 912 Angstroms" if this wavelength is the highest energy photon hydrogen can absorb.

So my question is: How can there exist a wide drop-out in the spectrum of galaxies at wavelengths shorter than about 91.2 nm, if at n = ∞, highest-energy electromagnetic radiation a hydrogen atom can emit or absorb is 91.1753 nm?

How is the hydrogen interacting with photons of higher-energies than this?

It must be a conceptual issue I'm not understanding.

Answer 1

The electron can scatter to $n=\infty + \epsilon$, by which I mean an unbound state.

The Lyman limit has an energy of:

$$ \frac{hc}{\lambda} = 13.59844\,{\rm eV} $$

which just happens to be the ionization energy of hydrogen. Of course, the density of final states is continuous, so all photons with greater energy will eventually be absorbed.