If $f(x)=x+\cos{x}\sin{x}$, and $g(x)=f(x)e^{\sin{x}}$,
then $L=\lim\limits_{x\to\infty}\frac{f'(x)}{g'(x)}=0$.
(This example is used to illustrate the fact that L'Hopital's rule requires that the derivative of the denominator is not zero in an interval around the value that is approached by $x$.)
My question is, does $L$ really equal $0$?
We have $\dfrac{f'(x)}{g'(x)}=\dfrac{2\cos^2{x}}{e^{\sin{x}}\cos{x}(x+\cos{x}\sin{x}+2\cos{x})}$.
For $x=(2k-1)\frac{\pi}{2}, k\in\mathbb{Z},$ we have that $\dfrac{f'(x)}{g'(x)}$ is undefined. So according to the epsilon-delta definition of limit (the section called "Finite limit at infinity"), $L$ does not exist.
According to the definition that you linked to, you're right. With that definition, the assumption that the denominator $g'$ is nonzero is implicit in the very definition of what it means for $f'(x)/g'(x)$ to have a limit, so it's not really an issue as far as l'Hospital's rule is concerned. (For the example you give, l'Hospital's rule would of course still not work, but for a different reason, namely that the limit of $f'(x)/g'(x)$ doesn't even exist in the first place.)
But many other sources use a more general definition, where it is only required that the domain of the function that you're taking the limit of contains arbitrarily large numbers (but not necessarily all sufficiently large numbers!), and that the epsilon stuff only holds for those $x$ that are actually in the domain. And with that definition, the example is correct, the limit is really $L=0$. (But l'Hospital's rule fails, since the assumption about the denominator is not fulfilled.)