Division by 0 in a rate of change problem This is not to ask for a solution, rather I want comments on the rigor of the step. Thank you for your time!

The graph shows a system of two points at $t$ and $t+dt$, it is a bit exaggerated of course.
The upper point is moving strictly in the $x$ direction and has a constant velocity $u$, while the lower point is always moving towards the upper point with constant velocity $v>u$. I want to find the rate of change of the distance $\mathcal{L}(t)$ which separates them (to find the time of collision $\tau$).
Keeping in mind that $\cos(\pi-\theta)=-\cos(\theta)$ and the cosines' law:
$$\begin{align*}
\mathcal{L}^2(t+\Delta t)&\approx(u\Delta t)^2+(\mathcal{L}(t)-v\Delta t)^2+2(u\Delta t)(\mathcal{L}(t)-v\Delta t)\cos(\theta(t))\\
&\approx(u\Delta t)^2+\mathcal{L}^2(t)-2v\mathcal{L}(t)\Delta t+(v\Delta t)^2+2u\mathcal{L}(t)\cos(\theta(t))\Delta t-2uv\cos(\theta(t))(\Delta t)^2
\end{align*}$$
$$\begin{align*}
\mathcal{L}^2(t+\Delta t)-\mathcal{L}^2(t)&\approx(u\Delta t)^2-2v\mathcal{L}(t)\Delta t+(v\Delta t)^2+2u\mathcal{L}(t)\cos(\theta(t))\Delta t-2uv\cos(\theta(t))(\Delta t)^2\\
\frac{\mathcal{L}^2(t+\Delta t)-\mathcal{L}^2(t)}{\Delta t}&\approx u^2\Delta t-2v\mathcal{L}(t)+v^2\Delta t+2u\mathcal{L}(t)\cos(\theta(t))-2uv\cos(\theta(t))\Delta t
\end{align*}$$
Now let $\Delta t\to0$, we have:
$$\begin{align*}
\frac{d}{dt}\mathcal{L}^2(t)&=-2v\mathcal{L}(t)+2u\mathcal{L}(t)\cos(\theta(t))\\
\Leftrightarrow 2\mathcal{L}(t)\mathcal{L}'(t)&=-2v\mathcal{L}(t)+2u\mathcal{L}(t)\cos(\theta(t))\\
\Leftrightarrow \mathcal{L}'(t)&=-v+u\cos(\theta(t))
\end{align*}$$
My problem is that when $t=\tau$, $\mathcal{L}(\tau)=0$ and so the division is  undefined. How can I go around this, or is the simplification valid (why)?
 A: Your argument is being applied at a time $t<\tau$. Also, since $\Delta t\to0 $, $ t+\Delta t<\tau$.
Your formula for $\mathcal{L'}(t)$ has  therefore been obtained for $t<\tau$ by division by a non-zero quantity. 
You are right to be concerned about this formula for $t=\tau$. It does not apply there and this is obvious if you make $u=0$ since then the formula gives the incorrect answer $-v$ instead of $0$.
A: At $\ t=\tau\ $, the derivative of $\ \mathcal{L}\ $ will not be defined, no matter how the points continue to move after that time. A left derivative, $\ \partial_-\mathcal{L}(\tau)=\lim_\limits{\delta\rightarrow0}\frac{0-\mathcal{L}(\tau-\delta)}{\delta}=u\cos(\theta(\tau))-v\ $ does exist, but this is strictly negative (not $0$, as you have supposed), because $\ v>u\ $.  Since $\ \mathcal{L}(\tau)=0\ $, and $\ \mathcal{L}(t)\ge0\ $ for any $\ t>\tau\ $ (if, in fact, defined), then if $\ \mathcal{L}\ $ has a right derivative at $\ t=\tau\ $ it must be non-negative, and therefore cannot be equal to the left derivative.
For the purposes of determining the value of $\ \tau\ $, none of this matters, because you only need to use $\ \mathcal{L}'(t)\ $ for $\ t<\tau\ $.
