Why isn't the terminal velocity completely achieved but only asymptotically achieved due to action of resistive force? Motion against resistive forces is given by: $$F_0 - R(v) = m\frac{dv}{dt}$$ where $R(v) = A.v + B.v^2$. Terminal velocity can be found by solving $$B.v^2 + A.v - F_0 = 0$$ ie. $v_t$ is the velocity when $\sum \mathbf{F} = 0$. But it is written by A.P. French that:

Note that the contrast between the sharply defined value of $v_t$ in the graph of $R(v)$ versus $v$, and the gradual manner in which the velocity is approached but never quite, in principle, reached.

Now, why can't it be reached?

 A: 
Now, why can't it be reached?

The short answer is that, due to the nature of the equation of motion, if the particle velocity $v$ is the terminal velocity $v_t$ at some time $t$, then $v = v_t$ for all time $t$.
That is to say, a solution to the equation of motion is 
$$v(t) = v_t$$
in which case, the acceleration is zero and thus, the velocity is unchanging with time.
It follows that if the velocity is not equal to $v_t$ at some time $t$, it's not equal to $v_t$ for any time $t$.
A: It's an aymptotic behavior. To understand the statement rigorously, you need to solve the differential equation to find velocity as a function of either time or distance travelled (to do the latter, write $\dot{v} = v\,\mathrm{d}_x\,v$ and you have a DE for $v$ as a function of $x$). However, physically, witness that as the object gets nearer and nearer to terminal velocity, the nett force on it gets nearer and nearer to nought (gravity becomes perfectly opposed by drag), so there is less and less tendency for the body's state of motion to change. 
This kind of situation arises everywhere throughout physics as stable systems approach their stable equilibrium points. That this is a general behavior can be understood by stability theory and the study of so called stable fixed points: near any equilibrium point differential equation systems can be linearized and linear DEs are solved by superpositions of functions of time of the form $e^{\alpha_j\,t}$: the system is stable iff all the $\alpha_j$ have negative real parts, so you get the classic $e^{-t}$ dependence of "distance" from equilibrium versus time. Compare the situation to where your cup of tea is cooling down: as it approaches room temperature, there is less and less temperature gradient "driving" the heat transfer, so in theory it never quite reaches room temperature.
A: Because it's only a limit for velocity, to which it will be continuously approaching by decreasing the difference between the actual velocity and this limit (and decreasing overall acceleration (pace of changing this velocity) toward zero).
