The problem
I have the following equation:
$$z=\frac{KAe^{\frac{1-e^{-At}}{A}}}{A+e^{\frac{1}{A}}\left(Ei({-\frac{1}{A}})-Ei(-\frac{e^{-At}}{A})\right)}$$ where $A>0$, $K>0$, $t>0$, and $Ei$ is the exponential integral function.
What I have found is that, when $A$ is small, z is well-approximated by
$$z^*=Ke^{-At}$$
The plot below shows this for $A=0.01$ and $K=1000$.
I am hoping someone can show formally why this approximation works so well?
My attempted solution to the problem
One idea I had was to solve $z=xz^*$ for $x$, then linearize the result around $a=0$.
This gives $x \approx 1$. Thus, when $a$ is nearly zero, $z \approx z^*$.
However, this "approximation" is clearly not optimal and I am curious if anyone has a better solution (if such a solution exists at all).
I am also curious if my solution counts as an approximation or just shows that $z$ and $z*$ are similar functions?
Considering $$\frac{z}K=\frac{Ae^{\frac{1-e^{-At}}{A}}}{A+e^{\frac{1}{A}}\left(\text{Ei}({-\frac{1}{A}})-\text{Ei}(-\frac{e^{-At}}{A})\right)}$$ expanding as series for small values of $A$ (assuming $A>0$ and $t>0$), we have $$\frac{z}K=1 + \left(1-t-e^{-t}\right)A+O\left(A^2\right)$$ while $$\frac{z^*}K=e^{-At}=1-t A+O\left(A^2\right)$$
More interesting is $$\frac{z}{z^*}=1+(1-e^{-t})A+O\left(A^2\right)$$ which tends to show that $z^*$ is a lower bound of $z$.
This is confirmed by your plot and others I made.