I'm asked specifically to use a Taylor series to determine the behavior of the integrand near the lower bound, $x=0$, to decide if the integral converges or diverges.
I used the Taylor series for $e^x$:
$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$
$$ e^{-x} = \frac{(-x)^0}{0!} + -x + H.O.T. = 1 + -x + H.O.T.$$
$$ \frac{e^{-x}}{x} = \frac{1}{x} + \frac{-x}{x} + H.O.T = \frac{1}{x} + \mathcal O\left(1\right)$$
Then focusing on the first behavior of the leading order term near 0, where the integrand blows up:
$$ \lim\limits_{x \to 0^- } \frac{1}{x} = \infty $$
If I've calculated correctly, then I believe the integral diverges. But my answer is not correct. What am I missing?
Your answer is absolutely correct, and that's a fine way to see it, at least intuitively. It would suffice also to note that $e^{-x}$ is bounded at $0$, so it won't help with the blowing up of $\frac1x$.
It's quick to prove otherwise though, note that $e^{-x}\geq e^{-1}$ on $[0,1]$, so for a $\delta>0$, $$ \int_\delta^1 \frac{e^{-x}}{x}\mathrm dx\geq e^{-1}\int_\delta^1\frac{1}{x}\mathrm dx $$ and taking limits $$ \lim_{\delta\to 0^+}\int_\delta^1 \frac{e^{-x}}{x}\mathrm dx\geq \lim_{\delta\to 0^+}e^{-1}\int_\delta^1\frac{1}{x}\mathrm dx=+\infty $$