Will the speed of object be the same for observers with different speeds if the speed of the object is closer to the speed of light? If we assume an object X traveling at a speed $V$ that is a bit  slower than the speed of light (let's say 1000 miles/sec.), what will be the difference in the observations of the speed of object made by two observers A & B, where:

*

*A is not moving at all with respect to the object under observation

*B is moving with a velocity $c-V$ in the direction of the moving object X, i.e., both are moving towards each other on the same imaginary line.

Please excuse if the question seems too lame for you.
 A: (Sorry for the awkward formatting - I'm new to the site.)
First, let's get rid of our privileged observer who gets to decide who is "really" traveling at what velocities. Instead, let's define our system in terms of relative velocities. If I understood your paragraph correctly, your system has:

*

*Let the velocity of $X$ measured by $A$ be $v_{AX}$


*Let the velocity of $B$ measured by $A$ be $v_{AB}$


*Let the velocity of $A$ measured by $B$ be $v_{BA}$


*Let the velocity of $X$ measured by $B$ be $v_{BX}$


*Given $v_{AX}$ and $v_{AB}$, solve for the other terms.
In the Newtonian paradigm, which approximates the relativistic paradigm for $v\ll c$, we first need to find $v_{BA}$. This is easy: it's just the negative of $v_{AB}$, since we switched our point of view to the opposite direction. Now we just add the vectors.
$$v_{BX}=v_{BA} + v_{AX} = v_{AX} - v_{AB}$$
In the relativistic paradigm, we need to account for the speed of light being the same for all observers. You can find one derivation here.
On a line, we can simply replace each vector with its matching scalar.
$$v_{BX} =  (v_{AX} - v_{AB})/(1 - v_{AX}\cdot v_{AB} / c^2)$$
In 3D space, we'd need to break each vector $v_{BX}$, $v_{AX}$, and $v_{AB}$ up into their $x$, $y$, and $z$ components, compute each separately, and then take their vector sum to obtain our vector $v_{BX}$.
As a side-note, 1000 miles per second (about 1600 km/s) isn't very close to the speed of light (about 300000 km/s).
A: $v_{XA}=0$. The value of $v_{XB}$ will be the result of the relativistic velocity addition of velocities $V$ and $c-V$, which is greater than $c-V$ and less than $c$.
