special relativity - What is $c + (-c)$?

If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is $$ v \oplus w = \frac{v+w}{1+vw} $$

As expected, $v \oplus 1 = 1$, as "nothing can go faster than light".
Similarly, $v \oplus -1 = -1$. (same thing in the other direction)

But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$ Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.

So what would the ground observer see? Is this even a meaningful question?

Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.