A limiting case
The expression "a → 0" is sloppy because a has units of inverse time squared, and the number 0 has no units. The proper statement is that Eq. (4.2) reduces to x(t) = C + Dt when at ≪ 1, or equivalently when t ≪ 1/a, which is now a comparison of quantities with the same units. The smaller a is, the larger t can be. Therefore, if "a → 0," then t can basically be anything. Assuming at ≪ 1, we can write
e±at ≈ 1 ± at, and Eq. (4.2) becomes
x(t) ≈ A(1 + at) + B(1 − at)
= (A + B) + a(A − B)t
≡ C + Dt. (4.63)
C is the initial position, and D is the speed of the particle. If these quantities are of order 1 in the units chosen, then if we solve for A and B, we see that they must be roughly negatives of each other, and both of order 1/a. So if the speed and initial position are of order 1, then A and B actually diverge in the "a → 0" limit. If a is small but nonzero, then t will eventually become large enough so that at ≪ 1 won't hold, in which case the linear form in Eq. (4.63) won't be valid.