Work so far:
$$f(x) = \frac{e^{-x}}{x}$$
$$e^{-x} = 1 - x + \frac{x^2}{2!} + \cdots+ \frac{(-1)^nx^n}{n!} + \cdots$$
$$\frac{1}{x} = 1- (x-1)+(x-1)^2 + \cdots + (-1)^n(x-1)^n + \cdots$$
$$\frac{e^{-x}}{x} \approx [1-x+\frac{x^2}{2!}][1-(x-1)+(x-1)^2] $$
Now I'm confused because I have multiplied the two expansions together (the first three terms of both expansions) to try and get the expansion for $f(x)$, but I am not sure if that is correct. I know I can do it by finding the derivatives of $f(x)$ and then just writing out the Maclaurin series, but I need to do it by using the parent functions and their expansions.
Using the Maclaurin series (Taylor series centered around $0$) of $e^x$: $$\frac{e^{-x}}{x}=\frac{1}{x}\sum_{k=0}^\infty\frac{(-x)^k}{k!}=\sum_{k=0}^\infty\frac{(-x)^k}{k!\cdot x}=\sum_{k=0}^\infty (-1)^k\cdot\frac{x^k}{k!\cdot x}=\sum_{k=0}^\infty (-1)^k\cdot\frac{x^{k-1}}{k!}$$ Which can be expanded into: $$\frac{1}{x}-1+\frac{x}{2}-\frac{x^2}{6}+\frac{x^3}{24}-\frac{x^4}{120}+\cdots$$ The series representation has the same boundaries as $\frac{e^{-x}}{x}$: it's defined for all $|x|>0$ and $x\not\to -\infty$.