The Intuition behind l'Hopitals Rule
The intuition is that although both numerator and denominator tend to zero or infinity, what eventually matters is their respective rate of change. They do not approach zero or infinity at the same rate and thus the one with the highest rate of change dominates the other.
There is a case where the result is obvious: Suppose that $f,g$ are continuously differentiable at $a$, that $g'(a)\ne0$, and $f(a)=g(a)=0$. Under these assumptions, $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}=\frac{f'(a)}{g'(a)}. $$ Of course, under the assumption that $f',g'$ are continuous at $a$, the latter is also $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$. Note that this is just a particular case of the result, but it is one of the most used cases, and it is widely applicable, for example, it also holds of analytic functions on an open domain in the plane, a case definitely not covered by the usual (real valued) argument.
The proof of the theorem is an extrapolation of this case, actually: We note that it is not necessary to assume that $f',g'$ are continuous at $a$, or even that they are defined there, since the differences $f(x)-f(a)$, $g(x)-g(a)$ are related to derivatives via the mean value theorem, so if $f(a)=0=g(a)$, we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\lim_{x\to a}\frac{f'(t_x)}{g'(t_x)} $$ for some $t_x$ between $x$ and $a$, so if $f'(t)/g'(t)$ converges as $t\to a$, then so does the displayed fraction.
The key is really the mean value theorem. Once one sees that this is all is needed in the case $f(a)=g(a)=0$, the other cases can be foreseen, and the argument adapted to cover them as well.
As you say, l'Hôpital's rule is due to Bernoulli, see here. You may also be interested in these slides by Ádám Besenyei on the history of the mean value theorem. Together with the history of the result, the geometric intuition discussed there may help you find the result more intuitive as well.
Around $x=a$ each of these functions can be approximated by their tangent line: $$ f(x)\approx f'(a)(x-a)+f(a) $$ $$ g(x)\approx g'(a)(x-a)+g(a). $$ Thus $$ \frac{f(x)}{g(x)}\approx \frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}. $$ Taking the limit of the right hand side gives $\frac{f'(a)}{g'(a)}$.