Suppose there are two funtions $f(x),g(x)$ such that (as $x \to a$) we have $f(x) \to +\infty$, $g(x) \to +\infty$, and $f'(x)/g'(x) = g(x)/f(x)$. Then by l'Hopital's rule, if $\lim f(x)/g(x)$ exists, then it is $1$.
How about a counterexample for the converse? I want $f'(x)=u(x)\cdot g(x)$, and $g'(x)=u(x)\cdot f(x)$ for some function $u(x)$, but $\lim f(x)/g(x)$ does not exist in $[-\infty,+\infty]$.
(The question was inspired by https://matheducators.stackexchange.com/questions/10427/ )
Given a function $u$, the system of differential equations $f'(x) = u(x) g(x)$, $g'(x) = u(x) f(x)$ has general solution $$ f(x) = A \exp(U(x)) + B \exp(-U(x)),\ g(x) = A \exp(U(x)) - B \exp(-U(x)) $$ where $U(x) = \int u(x)\; dx$. Thus $$ \dfrac{f(x)}{g(x)} = \dfrac{A \exp(U(x)) + B \exp(-U(x))}{A \exp(U(x)) - B \exp(-U(x)) } = \dfrac{A \exp(2U(x)) + B}{A \exp(2U(x)) - B }$$
In order to get an example where $\lim_{x \to a} f(x)/g(x)$ doesn't exist, we could take any nonzero $A,B$ and a function $U$ such that $\lim_{x \to a} U(x)$ doesn't exist.
Thus if $a = 0$, you might try $A=1$,$B=-1$, $U(x) = \sin(1/x)$, and $$\eqalign{f(x) &= \exp(\sin(1/x)) - \exp(-\sin(1/x))\cr g(x) &= \exp(\sin(1/x) + \exp(-\sin(1/x),\cr u(x) &= -\cos(1/x)/x^2}$$