I believe that the frequency for the rotating disc of radius $r$ can be anything below $1/r$ (or $c/r$ in SI units).
I think the picture you consider is flawed because you do not take into account the measurement of the speed by the observer on the disc. He would have to time the passing by 2 measuring roads at the previously known laboratory distance ($v = L / \Delta t$) — but the measured time will appear dilated with respect to laboratory frame! So, the Lorentz factors will cancel out in your formula.
While researching the Ehrenfest's paradox, I've found an interesting analogy between it and Bell's spaceship paradox on the internet: http://www.physicspages.com/2015/02/26/lorentz-contraction-in-a-rotating-disk-ehrenfests-and-bells-spaceship-paradoxes/. I believe, this might provide you some ideas.
Now please, bear with me for a minute, I have a story to tell. This is how I reached those conclusions, I find this quite entertaining.
From the point of view of an observer riding the disc, disc's geometry is unchanged although a very strong force points towards the rim of the disc. Basically, he feels gravitational potential $U(r) = -\frac12 r^2 \omega^2$ where $\omega$ is the laboratory frame rotation frequency (see https://physics.stackexchange.com/a/53771/119172 and https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html for details):
$$ ds^2 = -(1 - r^2 \omega^2) dt^2 + dr^2 + r^2 d\phi^2 + dz^2 $$
A cool feature of this situation is that here, like in the case of the Schwarzschild black hole, we have an apparent singularity at the distance $r = 1 / \omega$. And if you compute the scalar curvature in this metric, you will get
$$ R = -\frac{2 \omega ^2 \left(r^2 \omega^2 -2\right)}{\left(r^2 \omega ^2-1\right)^2} $$
We see that at the curvature is singular at $r=1 / \omega$ providing us a physical singularity unlike the one in Schwarzschild's case.
What does it mean? As the observer does not known a thing about him being rotating, he can interpret the pull towards the rim as if something massive was surrounding the disc (although symmetric mass distribution should have zero net force). And at some distance this force goes through the roof as if there was a gorgeous cylindrical black hole!
On the other hand, lets go back to the laboratory frame. Say, we have this disc of the radius $r$ and we want to spin it up to frequency $\omega = 1 / r$. What linear speed will correspond to that? Well, $v = \omega r = 1$. So, the rim of the disc would have to move with the speed of light!
