Disclaimer: this is probably just a giant misunderstanding of implicit differentiation, but I'm not sure where!
I'm looking at an initial-value problem for a simple damped harmonic oscillator but where damping is proportional to the square of the velocity instead: $$ m\ddot{x} + kx + b\dot{x}^2 = 0 $$ for the initial conditions: $$x(0) = 0$$ $$\dot{x}(0) = v $$ where $v>0$ is the initial velocity and only from $t=0$ until $\dot{x} = 0$ (to ensure damping always points in the correct direction opposing velocity, as in the above formula!).
In particular, I'm interested in the maximum/minimum values of the acceleration $\ddot{x}$, which will occur when $\dddot{x} = 0$.
So, if we differentiate again with respect to time $t$, we obtain $$ m\dddot{x} + k\dot{x} + 2b\dot{x}\ddot{x} = 0 $$ and thus when $\dddot{x} = 0$, it will follow that $$ k\dot{x} = -2b\dot{x}\ddot{x} \Rightarrow \boxed{\ddot{x} = -\frac{k}{2b}}, \text{when }\dot{x} \ne 0$$
But, this makes no sense! It suggests that $\min\ddot{x}\to -\infty$ as $b\to 0$ (as in the undamped case).
So, there must be a mistake (or misunderstanding) here, but I can't seem to determine where!
*Edit: when $b \ne 0$, all minima/maxima values of $\ddot{x}$ occur when $\dot{x}\ne 0$ anyway, so this isn't a division-by-zero thing either...
*Edit #2: the above statement that $\dot{x} \ne 0$ at extreme values of $\ddot{x}$ may not be so true after all -- the numerical methods I use for IVP's to $$ m\ddot{x} + kx + b\dot{x}^2 = 0 $$ only list very small differences between the times when $\dot{x} = 0$ and when $\ddot{x}$ is at a maximum/minimum. So, it may actually be the case that every extreme value of $\ddot{x}$ occurs when $\dot{x} = 0$, and that my numerical solutions are simply just way too imprecise (I'm using WinPlot) to identify whether there is a real difference or not! I'm also thinking this is case because my math here appears to be correct...
Your solution makes sense. The polynomial you are solving at the time of optimality $t_*$ is,
$$\dot{x}(t_*) \left(k + 2\,b\,\ddot{x}(t_*)\right) = 0,$$
which is a (mixed) second-order polynomial in the unknowns. There are two possible solutions which you have identified. One is $\dot{x}(t_*) = 0$ and the other is when $\ddot{x}(t_*) = \frac{-k}{2\,b}.$ What happens when $b\to 0$? Well, the former polynomial decays to the linear equation,
$$k\,\dot{x}(t_*) = 0,$$
which only has one solution which is the former of the two presented solutions. It is not true that all solutions of the former polynomial converge to solutions of the latter. Therefore, the problem is with the assumption that the given root of the first polynomial tends to the solution of the second [linear] polynomial. So overall your math is right, but one cannot generalize that particular solution for the damped case into the undamped case.
It is also worth noting that if all optimal points occur at $\dot{x}(t_*) = 0$, you can't say anything about the value of $\ddot{x}(t_*)$ since it need not be what you found. I think you noted this in your question already.