Why the velocity $v$ is taken as value and not as definition in special relativity equations?

Why the velocity $$v$$ is taken as a value and the definition of velocity not applied on a relativistic equations?

The equations of time dilation and length contractions as we know are
$$L = {L_0}{\left( {1 - {{{v^2}} \over {{c^2}}}} \right)^{{1 \over 2}}}$$ and
$$\Delta t = {{\Delta {t_0}} \over {{{\left( {1 - {{{v^2}} \over {{c^2}}}} \right)}^{{1 \over 2}}}}}$$
Here, $${\Delta {t_0}}$$ is time interval between two events, and $${L_0}$$ is length in rest frame. An observer moving with velocity $$v$$ measures the time interval $$\Delta t$$ and the length $$L$$.

The velocity $$v$$ that I am using has some value and for some reason I take it as a value and do not apply the definition
$$v = {{dx} \over {dt}} = \mathop {\lim }\limits_{\Delta t \to 0} {{\Delta x} \over {\Delta t}} = \mathop {\lim }\limits_{t \to {t_0}} {{x - {x_0}} \over {t - {t_0}}}$$
Velocity as I know is "change in displacement over change in time". Let's take length contraction equation, as the length contract there must be some variation for the end points of a rod (say) to move across a point, causing change in $$v$$. So, why I am not taking some relativistic equation of $$v$$ here in the equation. I feel, somehow there is some connection between length contraction, time dilation and this velocity. There must be something in this $$v$$ which is not factored in these equations.

• Would the fact that $v$ represents the velocity of the moving frame (relative to a frame considered stationary) help see why it's a constant value? – Kyle Kanos Jul 5 '19 at 14:22
• @kyle the moving frame will have contraction, and the question remains the same. – Roopesh Singh Jul 5 '19 at 16:37
• The frame doesn't have a contraction, but objects within the two frames will have different lengths based on the relative speed of the moving vs stationary frame. – Kyle Kanos Jul 5 '19 at 18:10
• Even though a stationary observer will see an object as contracted, the front of that object will pass by him with the same velocity $v$ as the rear of that object. – Bill Watts Jul 5 '19 at 22:55
• @KyleKanos Why not? Isn't frame all about space-time? – Roopesh Singh Jul 8 '19 at 13:17