Not sure if I should directly ask in the comment section of that thread or start a new thread like I have now. But anyway, I happen to come across this thread on MSE:
Asymptotic Inner and Outer Expansion for a Function
and it deals with the inner and outer expansion of the function $$f(x,\epsilon) = \frac{e^{-x/\epsilon}}{x}+\frac{\sin (x)}{x}-\coth (x)$$
where $x>0$ and $0<\epsilon \ll 1$.
After trying this problem for myself, I found the outer region to be $x=O(1)$ with expansion $$f(x,\epsilon) \sim \frac{\sin (x)}{x} - \coth (x)$$
On the other hand, the inner region is $x=O(\epsilon)$, but when I tried to compute the inner expansion via the rescaling $x=\epsilon y$ where $y=O(1)$ as $\epsilon \rightarrow 0$, I got
\begin{align} \ f(x,\epsilon) & =F(y,\epsilon) \\ \ & = \frac{e^{-y}}{\epsilon y} + \frac{\sin (\epsilon y)}{\epsilon y} - \coth (\epsilon y) \\ \ & = \Bigl (\frac{e^{-y}}{y}\epsilon ^{-1} \Bigr)+ \Bigl(1-\frac{(\epsilon y)^2}{3!}+\frac{(\epsilon y)^4}{5!}+\cdots \Bigr)- \Bigl((\epsilon y)-\frac{(\epsilon y)^3}{3}+\frac{2(\epsilon y)^3}{15}-\cdots \Bigr)^{-1} \\ \end{align}
and we get a $(\frac{e^{-y}}{y}-\frac 1y) \epsilon ^{-1}$ term in the expansion even though we have assumed that $y = O(1)$.
How do I get around this problem? Thanks!