Converse of L'Hospital

Question

L'Hospital's Rule: If ${f(x)}\over{g(x)}$ is either $0\over0$ or $\infty\over\infty$, then the $\lim_{x\to \infty}$${f(x)}\over{g(x)}$ $=$ $\lim_{x\to \infty}$${f'(x)}\over {g'(x)}$. What I want to know is if the converse is true.

In other words, if $\lim_{x\to \infty}$${f(x)}\over{g(x)}$ $=$ $\lim_{x\to \infty}$${f'(x)}\over {g'(x)}$, then is ${f(x)}\over{g(x)}$equal to one of $0\over0$ or $\infty\over\infty$? By this I mean that there does not exist a case in which L'Hopital does not apply yet gives the correct answer.
It is not a repeat of another question, as many have suggested. I would ask those people to take a look at the question that they have said I have repeated. These are two different questions, and it surprises me that they could have been confused.

Answer 1

Notice that $$\lim_{x \to \infty} \frac{1}{x^2} = \lim_{x \to \infty} \frac{0}{2x} = 0$$ even though the initial limit did not have an indeterminate form. That is, it is possible that you could get the right answer by applying L'Hopital even in cases where it is invalid. You could think of this as the calculus equivalent of the fake proof that $$ \frac{16}{64}=\frac{1}{4} $$ where you cancel out the sixes.

Answer 2

Let $f(x)=x+\sin(x), g(x)=x$.

Then $$\lim_{x \to \infty} \frac{f(x)}{g(x)} =2$$ by the squeeze theorem but $$\lim_{x \to \infty} \frac{f'(x)}{g'(x)} =\lim_{x \to \infty} 1+\cos(x) = DNE$$