What is the definition of $f$ falling slower than $g?$ The German Wikipedia is full of articles like  this one saying one function falls slower than another. I wrecked google, but could not find the definition or sufficient conditions. Is it:
$ \lim_\limits{x \to \alpha }\frac{f(x)}{g(x)} = 0$, or
$\displaystyle{\frac{d}{dx}\frac{f(x)}{g(x)} < 0} $ (in appropriate region)
Or is any of these sufficient? I guess the first one, since it doesn't require differentiability.
 A: Let's restrict this discussion to those stretched exponentials in the article you linked to, since they have the property that their limit is zero as $t \to +\infty$, which makes talking about the concept of "falling" relatively straightforward.
If we have two such functions, $f$ and $g$, then to say "$g$ falls more slowly than $f$" means that $g$ approaches $0$ more slowly than $f$ does.  In other words, $f$ falls (decreases) more quickly than $g$ does.  So it is the case that $\displaystyle\lim_{t\to +\infty} \frac{f(t)}{g(t)} = 0$, because the numerator approaches $0$ more quickly than the denominator does.  There are two ways of stating this here:
$$ \lim_{t \to+\infty} \frac{f(t)}{g(t)} = 0 \qquad \text{and} \qquad \lim_{t\to+\infty} \frac{g(t)}{f(t)} = +\infty$$
In general, $g(t) = e^{-t^\beta}$ falls more slowly than $f(t) = e^{-t}$ if we have $\beta < 1$.  That's just from the link you provided.  Consider:
$$ \lim_{t\to +\infty} \frac{g(t)}{f(t)} = \lim_{t\to +\infty} \frac{e^{-t^\beta}}{e^{-t}} = \lim_{t\to+\infty} \frac{e^{-t^\beta} \cdot e^t}{e^{-t} \cdot e^t} = \lim_{t\to+\infty} e^{t-t^\beta}$$
Since $\beta < 1$ and $t \to +\infty$, then $t^\beta \to 0$.  Thus $t - t^\beta \to +\infty$ as $t \to +\infty$, and so $\displaystyle \lim_{t\to+\infty} e^{t-t^\beta} = +\infty$.  This paragraph isn't exactly rigorous, but I'm going for clarity here.
A similar argument shows the other limit mentioned above: $\displaystyle\lim_{t\to+\infty} \frac{f(t)}{g(t)} = 0$
