Past exam of the short options course of the University of Oxford / 2023 / 8 / b / iii / Solution

When $0\le v,v^{\prime }\ll c$, expand the denominator:The leading term is ordinary Galilean velocity addition. When both $v,v^{\prime }\to c$ from below, the exact formula instead gives $\boxed{v^{\prime \prime }\to c}$, rather than $2c$. Indeed, writing $a=v/c$, $b=v^{\prime }/c$, we have $1-v^{\prime \prime }/c=(1-a)(1-b)/(1+ab)>0$ whenever $a,b<1$.