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 $?
It is important to think about what the velocity addition formula is meant to do. The velocity addition formula should be applied in situations of this type:
When you (person $A$) see someone (person $B$) speeding by you with velocity $v$, and when this second person sees another person (person $C$, one can also substitute objects for people, of course) pass with velocity $u'$, then what is the velocity $u$ at which person $A$ sees person $C$ come by?
As @WillO already pointed out in the comments, if person $B$ is speeding by at $c$, then there is no reference frame in which he/she/it is stationary. Therefore, it is clear that the velocity addition formula should not be applied here. Note that this only rules out situations where both $|v|$ and $|u'|$ equal $c$, since the formula is symmetrical under interchange of the two.
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.
