The problem we have is $x'' = ax$. When $a=0$ then the solution is $x=x_0 + v_0 t$. We're interested in showing that the solution (when $a \neq 0$) as $a\to 0$, tends to the solution we get with $a=0$: $$\lim_{a\to 0}x(t;\ a)=x_0 +v_0t$$ When $a<0$ we have the equation of a harmonic oscillator and the solution is $$x_-(t;\ a)=c_1 \cos(\sqrt{a}t) +c_2\sin(\sqrt{a}t)$$
When $a>0$ we don't have the equation of a harmonic oscillator, the solution is $$x_+(t;a)=c_1 e^{ \sqrt{a}t}+c_2 e^{-\sqrt{a}t}$$
So we want to show both these solutions tend to $x_0+v_0t$ when $a\to 0$. How do we deal with such a problem? We can't simply take $a=0$, as it's both naive and it yields a result different than the one we want.
For $x_+$ I thought the following: when $a\to 0$ then the two solutions tend to become the same, but the general solution has to be a linear combination of two independent solutions, therefore as $a\to 0$ the solution will be $$x(t;\ a\to 0^{+})=c_1 e^{\sqrt{a}t} +c_2te^{\sqrt{a}t} $$ which taking $a=0$ tends to the one we want. However I don't know if it's rigorous enough, plus I can't seem to do anything with $x_-$. Thanks in advance :)