What is wrong with my logic and derivatives? Since the derivative of a function is analogous to a description on how fast the function grows, I thought that $\lim_{x\to\infty}\frac{f'(x)}{f(x)}=0$ for the following reason.
Assume $f(x)$ is monotone increasing on an interval $(a,\infty]$.  If $f'(x)$ is large, that means that $f(x)$ is getting larger and larger.  From this, I've assumed that $f'(x)$ must eventually be outdone by $f(x)$ simply because $f'(x)$ cannot get large without making $f(x)$ even larger.
But that's obviously not the case because if $f(x)=\Gamma(x)$, then $\lim_{x\to\infty}\frac{f'(x)}{f(x)}\ne0$.
I'm just wondering what is wrong with my logic, that's all.
 A: In short there exist functions such as $e^x$ whose derivative increases at a rate equal to the function itself. You make an incorrect assumption precisely here:

"From this, I've assumed that $f′(x)$ must eventually be outdone by $f(x)$ simply because $f′(x)$ cannot get large without making $f(x)$ even larger."

With that being said, these sorts of questions are really quite laudable.
A: Just because $f(x)$ is "getting larger and larger" doesn't mean that $f'(x)$ "must eventually be outdone by $f(x)$." For example: what if $f'(x)$ itself gets larger and larger, at the same or greater rate than $f(x)$, and starts out equal to or greater than $f(x)$? The comments and other answer give you simple counterexamples. The function $x\mapsto e^x$ has this property: its derivative starts out at the same value and grows at the same rate.
Moreover, I suspect your intuition about asymptotic growth at infinity is wrong: even if the claim above were true, it would not entail that $\lim_{x\to\infty}\frac{f'(x)}{f(x)}=0$. 
In other words: just because $g(x)$ is eventually larger than $h(x)$ (that is, $g>h$ for all sufficiently large $x$) does not mean their ratio $h/g$ tends to 0 as $x$ tends to infinity. For example, the function $g(x)=2x$ is eventually greater than the function $h(x)=x$, but that does not mean $h/g=\frac{1}{2}$ tends to zero at infinity.
