Inner and outer expansions I'd really appreciate some help with the following question.

Find inner and outer expansions, correct up to and including terms of O(ε), for the function


$$ f(x;ε) = \frac{e^{-\frac{x}{ε}}}{x} + \frac{\sin x}{x} - \coth x $$

By setting $x=O(1)$, the first term becomes exponentially small, and so (I think) the outer expansion is simply
$$ f(x;ε) = \frac{\sin x}{x} - \coth x $$
To find the inner expansion, I take $x=O(ε)$. I therefore rescale $x$ using $x=εX$, with $X=O(1)$, and set $f(x;ε)=F(X;ε)$.
Plugging this into the original equation, I obtain
$$ F(X;ε) = \frac{e^{-X}}{εX} + \frac{\sin(εX)}{εX} - \coth(εX) $$
I'm not sure if this is right and, if so, how to continue? At some point, I may need to use matching?
Thanks in advance for any help!
Edit (following Lutz's comment):
Using Taylor expansions, we can write
$$ \frac{\sin(εX)}{εX} ≈ 1 - \frac{1}{6}ε^2X^2 $$
$$ \coth(εX) ≈ \frac{1}{εX} + \frac{εX}{3} $$
$$ \frac{e^{-X}}{εX} ≈ \frac{1}{εX} - \frac{1}{ε} + \frac{X}{2ε} - \frac{X^2}{6ε} $$
Then plugging into our original equation, the $\frac{1}{εX}$ terms cancel, and we are left with:
$$ F(X;ε) = 1 - \frac{1}{6}ε^2X^2 - \frac{εX}{3} - \frac{1}{ε} + \frac{X}{2ε} - \frac{X^2}{6ε} + ... $$
I assume this isn't the final answer, but I'm unsure of how to rewrite the last terms, which have $ε$ in their denominators?
Thanks again!

Now insert that for $x\approx 0$ you have
$$\frac{\sin x}{x}\approx 1-\frac16x^2$$
and
$$
\coth(x)=\frac{e^x+e^{-x}}{e^x-e^{-x}}\approx\frac{1+\frac12x^2}{x+\frac16x^3}\approx \frac1x+\frac13x
$$
to cancel out the singularities at $x=0$.
