Singular perturbation of a first order linear differential equation In studying a reflection-transmission problem involving exotic materials, I have come across the following linear first-order differential equation:
\begin{equation}\tag{1}\label{eq:1}
A\frac{\partial}{\partial t}g(t) + Bg(t) = f(t),
\end{equation}
where $A$ and $B$ are constants, $g(t)$ is associated with the reflected wave, and $f(t)$ is a (finite) driving function associated with the incident wave.  Both $A$ and $B$ may be positive or negative.  I am interested in the behavior of the solution in the limit that $A\rightarrow0$.
I know there is an exact solution to Eq. \eqref{eq:1}, which is
\begin{equation}
g(t) = C e^{-Bt/A} + \frac{1}{A}\int_{-\infty}^t e^{-B(t-t')/A}f(t')dt',
\end{equation}
where $C=0$ because $g(t)=0$ if $f(t)=0$.  However, I do not understand how this exact solution reduces to the case where $A=0$, which is $g(t)=B^{-1}f(t)$.  Any insight would be greatly appreciated.
I've seen a lot of documents discussing asymptotic analyses of linear differential equations (for example, see these lecture notes), but they all start with second-order equations.  Is this because there is inherently problematic with first-order?  

I presume $A > 0$ and $B > 0$.   A change of variables $s = (t-t')/A$
in the integral gives you 
$$ g(t) = C e^{-Bt/A} + \int_0^\infty e^{-Bs} f(t - A s) \; ds $$
Now as $A \to 0+$, $f(t-As) \to f(t)$ if $f$ is continuous.  Assuming $f$ is bounded, we can use the Dominated Convergence Theorem and this integral goes to 
$$f(t) \int_0^\infty e^{-Bs}\; ds = B^{-1} f(t)$$
