special relativity - A curious case of Relativistic Velocity Addition

The relativistic velocity addition formula is

$$u = \frac{v+u'}{1+ \frac{vu'}{c^2}}$$

Where

$u$ = velocity of projectile seen by rest observer "A"

$v$ = velocity of moving observer "B" as seen by rest observer "A"

$u'$ = velocity of projectile seen by B

Now the question is this: If $v=c,$ and $u'=-c $

I get an undefined answer. i.e. the relavistic velocity addition formula is undefined.

Whats wrong with setting $v=c,$ and $u'=-c $?

Answer

This isn't definitive, since the answer is always undefined, but let's be cutesy. Let's let $u' = a*c$ and $v = -a*c$, where $a < 1$

Then,

$$\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{ac-ac}{1- \frac{a^{2}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c-c}{-2a}\\ &= 0 \end{align}$$

Now, the reason why this isn't definitive is that you can take different limits, if you want. Say, let $u' = a^{2}c$ and $v = -ac$

Then,

$$\begin{align} u &= \lim_{a\rightarrow 1}\frac{v+u'}{1+ \frac{vu'}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{a^{2}c-ac}{1+ \frac{a^{3}c^{2}}{c^2}}\\ &= \lim_{a\rightarrow 1}\frac{c\left(2a -1\right)}{3a^{2}}\\ &= \frac{c}{3} \end{align}$$

So, it's clear that, by taking the limit in different ways, you can get an arbitrary answer. It's not valid to choose an observer moving at the speed of light and then take velocities relative to that observer.