V=W THEN IT BECOMES (V-V)/(1-(V/C)^2)=0 (as v/c is not equal to 1 for them to cancel out the denominator)

@Sajith: I would interpret the result 0/0 as the question of "what's the speed of another object while I'm traveling at the speed of light" doesn't make sense in special realitivity. Why? When you measure the speed of another object (its "relative speed") you need to reference it against your "base speed" which you set to zero. If you travel at the speed of light you would need to make the speed of light your reference speed and set it to zero. By setting the speed of light to zero you are breaking the basic law of special relativity: "the speed of light is always a constant ~300m m/s for every observer". If you ignore this rule you're not working with the theory of special relativity, but a new theory. Good luck proving it with real world experiments! :) I found this article very helpful https://wtamu.edu/~cbaird/sq/2014/11/03/why-is-time-frozen-from-lights-perspective/

One still gets C, if we are careful enough to consider the limits properly. At first, consider V=C, and W=W itself. Then (V-W)/(1-VW/C^2) becomes C(C-W)/(C-W), which is simply C. This is irrespective of the value of W, i.e. even if W tends to C, we still have (C-W)/(C-W) = 1 as a limit.

I was amused to notice that the formula yields a maximum velocity of C even when either V or W exceeds C (which, as we have learned, is an impossibility). Example: if (impossibly) an object traveling at twice the speed of light, i.e. velocity 2C, would meet a light beam traveling straight towards it, the speed at with the light would hit that impossible object, remains C. The calculation is: (2C+C) / (1 + 2CxC/C^2) = 3C/3 = C.

Which is what you would expect. Relativistic effects only kick in when frames of reference are moving with respect to each other. If V=W the associated frames of reference are not moving with respect to each other and so we we should see no relativistic effects.