Present value heuristics problem
Intuition

Well, when $r$ is zero, you are just getting value $A$ per time unit, so total present value will be $A \cdot t$.



The approximation

You are asking about the approximation
$$
\frac A {-r}(e^{-rt}-1) \approx A \cdot t.
$$
This is equivalent to
$$
\frac 1{rt}(1-e^{-rt}) \approx 1,
$$
which is true for small $rt$ values. You can see this by taking the limit of the left hand side and using L'Hospital's rule.
$$
\lim_{rt \to 0} \frac{\frac{\text{d} (1 - e^{-rt})}{\text{d} rt}}{\frac{\text{d} rt}{\text{d} rt}}
= \lim_{rt \to 0} \frac{e^{-rt}}{1} = 1.
$$



The approximation in another form

The approximation
$$
1-e^{-h} \approx h
$$
seems to be a variant of the more well known approximation
$$
\ln(1+h) \approx h
$$
for small values of $h$.



How good is this approximation

To get a sense of the error term for large values using your approximation, you can use Taylor-approximation for the function
$$
f(x) = \frac{1-e^{-x}}{x}
$$
starting from some very small $x$ value.