Why doesn't a singularity spin infinitely fast? A question for a random reader passing through: because of the nature of things spinning faster when it gets smaller, wouldn't a singularity spin infinitely fast as it's infinitely small, if it has a spinn from the beginning that is? Btw I know what a "ringularity" but does it automaticly become one when rotational velocity is added? Is it just "fake news"?
Your intuition is correct, as far as classical dynamics is concerned. The main problem, I think, arises in that you are referring to a strictly quantum mechanical phenomenon - what is the physical singularity? In General Relativity (GR), we know these singularities exist, but we cannot really know more than that (aside from events that occur not too close to the physical singularity). But, it's deeper than that, because in GR we cannot get information out of a black hole, so we'd never really know. That's why people in the comments are referring to a "theory of quantum gravity" which would take over where GR breaks down (at the physical singularity). 

For a non-rotating (Schwarzschild) BH, the physical singularity is at $r = 0$, but the coordinate singularity is at the event horizon, $r = 2M$. 

For a rotating (Kerr) BH, the physical singularity is now a ring with no thickness but non-zero radius - this is the ringularity. We literally don't know what should happen here (need "quantum gravity"), but in GR we can't even know in principle, because if someone were to try to fall into the ringularity we would never receive any signals back from them (as far as GR tells us).

OKay, with all that out of the way, we can still make a similar comment like what you want to within the framework of GR:

Consider a Kerr BH (in Boyer-Lindquist coordinates) with dimensionless spin $a$ whose symmetry axis is along the $z$, so that the ringularity lies in the $x$-$y$ plane. One can show that at $z = 0$ and at $r = 0$ (where $r$ is not the physical radial distance), the equation of the ringularity is

$$ x^2 + y^2 = a^2$$

and so the radius of the ringularity is the spin. Therefore, as the spin of the Kerr BH increases, the radius of the ringularity also increases. This is the opposite of what your classical intuition leads to, but then again, this is a very nontrivial spacetime metric, and it exists on the fringe between classical and quantum physics.


  Why doesn't a singularity spin infinitely fast?


the ringularity is not spinning itself, rather it is defined by the spin of the BH. Mathematically, we can construct some kind of "spin" of the ringularity, but it wouldn't make much physical sense. 

Now, if the BH was spinning infinitely fast, i.e. $a \rightarrow \infty$, then the ringularity would have an infinite radius!!! And THAT's certainly problematic (but interesting!). Fortunately, as far as we know, BH's cannot have a spin greater than 1 - the maximum spin of a Kerr BH is capped by the naked singularity hypothesis. You see, if BH's could have a spin greater than 1, then the ringularity would extend "outside" the event horizon, which is problematic because it would mean that we could observe the collapse of an object into infinite density. ;)