Are there at least 45 variants of l'Hopital's Rule and how to prove variant with $\lim\limits_{x\to a^+} f(x)= \lim\limits_{x\to a^+} g(x) = \infty$? The following is a problem in Ch. 11 "The Significance of the Derivative" from Spivak's Calculus



*To complete the orgy of variations on l'Hopital's Rule, use Problem 55 to prove a few more cases of the following general
statement (there are so many possibilities that you should select just
a few, if any, that interest you):

If $\lim\limits_{x\to [\ ]} f(x)= \lim\limits_{x\to [\ ]} g(x) = \{\
 \}$ and $\lim\limits_{x\to [\ ]} \frac{f'(x)}{g'(x)}=(\ )$, then
$\lim\limits_{x\to [\ ]}  \frac{f(x)}{g(x)}=(\ )$. Here $[\ ]$ can be
$a$ or $a^+$ or $a^-$ or $\infty$ or $-\infty$, and $\{\ \}$ can be
$0$ or $\infty$ or $-\infty$, and $(\ )$ can be $l$ or $\infty$ or
$-\infty$.

First of all, given the problem statement above, is it accurate to say that there are at least $5\cdot 3 \cdot 3=45$ variations of L'Hopital's Rule?
I've seen questions on proving L'Hopital's Rule but given that there are many variations on the rule, and the fact that the proofs are not all exactly the same (I've gone through at least ten so far), I'd like to ask about one particular variation that I can't seem to find a proof for.
When we let $[\ ]$ be $a$, $a^+$, or $a^-$ and let $\{\ \}$ be $\infty$, for example, I can't seem to come up with a proof.
To cut to the chase, I'd like to know how to prove the following

If $\lim\limits_{x\to a^+} f(x)= \lim\limits_{x\to a^+} g(x) = \infty$ and $\lim\limits_{x\to a^+} \frac{f'(x)}{g'(x)}=l$, then
$\lim\limits_{x\to a^+} \frac{f(x)}{g(x)}=l$.

The proofs of l'Hopital's Rule I have seen usually start by using the assumption that the limit $\lim\limits_{x\to [\ ]} \frac{f'(x)}{g'(x)}=(\ )$ exists to establish existence of $f'$ and $g'$ and $g'\neq 0$ on a certain interval $A$ that depends on what the limiting value is in the limits, ie what is chosen as the $[\ ]$ parameter.
Next the Mean Value Theorem is invoked to establish either that $g(x)\neq 0$ or $g(x)-g(a)\neq 0$ on interval $A$.
Then the Cauchy-Schwarz MVT is invoked to obtain some relationship of one of the forms
$$\frac{f(x)}{g(x)}=\frac{f'(\alpha_x)}{g'(\alpha_x)}$$
$$\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\alpha_x)}{g'(\alpha_x)}$$
where $\alpha_x \in A$. At this point the limit of both sides is taken and we end up with the conclusion of the particular variant L'Hopital's Rule we are proving.
Obviously I skipped over a few steps in this description, but this is a general outline. I believe the issue I have for proving the statement I asked about above is application of MVT. If the limit of $f$ and $g$ when $x$ approaches $a$ is $\infty$, we can't use point $a$ to apply the MVT as is done in the proofs of all the other variations of L'Hopital's Rule as far as I can tell.
 A: Assumptions: $$\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^+}g(x)=\infty\qquad (1)$$ and $$\lim_{x\rightarrow a^+}\frac{f’(x)}{g’(x)}=l\qquad (2)$$
Assertion: $$\lim_{x\rightarrow a^+}\frac{f(x)}{g(x)}=l$$
To prove the assertion, one needs to show that $\forall \epsilon>0,\exists \delta>0$ such that $$x\in (a,a+\delta)\Rightarrow \left|\frac{f(x)}{g(x)}-l\right|<\epsilon.$$
Proof of Assertion. Let $\epsilon>0$ be given. By (2), there exists $\delta_1>0$ such that $$x\in (a,a+\delta_1)\Rightarrow \left|\frac{f’(x)}{g’(x)}-l\right|<\frac{\epsilon}2.$$
Let $$\epsilon_1=\min\left(\frac 12,\frac{\epsilon}{4(|l|+1)}\right).\qquad (3)$$
By (1), there exists $\delta$ with $0<\delta<\delta_1$ such that for $x\in (a,a+\delta),$ one has $\epsilon_{1,f}(x):=f(a+\delta_1)/g(x)$ satisfies $|\epsilon_{1,f}(x)|<\epsilon_1$ and $\epsilon_{1,g}(x):=g(a+\delta_1)/g(x)$ satisfies $|\epsilon_{1,g}(x)|<\epsilon_1.$
Now for the above $\delta>0$ and for $x\in (a,a+\delta)\subseteq (a,a+\delta_1),$ one has by Cauchy’s MVT that $$\frac{f’(\alpha_x)}{g’(\alpha_x)}=\frac{f(x)-f(a+\delta_1)}{g(x)-g(a+\delta_1)}=\frac{\frac{f(x)}{g(x)}-\epsilon_{1,f}(x)}{1-\epsilon_{1,g}(x)},$$ where $\alpha_x\in (x,a+\delta_1).$ It follows that $$\left|\frac{f’(\alpha_x)}{g’(\alpha_x)}-l\right|=\left|\frac{\frac{f(x)}{g(x)}-\epsilon_{1,f}(x)}{1-\epsilon_{1,g}(x)}-l\right|<\frac {\epsilon}2$$
$$\Rightarrow \left|\frac{f(x)}{g(x)}-\epsilon_{1,f}(x)-l(1-\epsilon_{1,g}(x))\right|<\frac {\epsilon}2(1-\epsilon_{1,g}(x))$$
$$\Rightarrow \left|\frac{f(x)}{g(x)}-l\right|-\left|l\epsilon_{1,g}(x)-\epsilon_{1,f}(x)\right|<\frac{\epsilon}2(1+\epsilon_1)$$
$$\Rightarrow \left|\frac{f(x)}{g(x)}-l\right|<\frac{\epsilon}2(1+\epsilon_1)+(|l|+1)\epsilon_1\leq \frac {\epsilon}2\left(1+\frac 12\right)+\frac{\epsilon}4=\epsilon,$$ where the last estimate follows from (3). This proves the assertion. QED
A: There are not so many variants of l’Hôpital’s theorem: just $0/0$ and $\infty/\infty$.
You can simply prove them for the case
$$
\lim_{x\to a^+}\frac{f(x)}{g(x)}
$$
because the case for $x\to a^-$ follows by replacing $x\mapsto -x$ and the case $x\to\infty$ follows by replacing $x\mapsto 1/x$. The case $x\to a$ combines the two above.
Indeed, if we examine the last case, we have to do
$$
\lim_{x\to0^+}\frac{f(1/x)}{g(1/x)}
$$
and it's simple to verify that the hypotheses are satisfied, so we want to compute
$$
\lim_{x\to0^+}\frac{-f'(1/x)/x^2}{-g'(1/x)/x^2}=\lim_{x\to0^+}\frac{f'(1/x)}{g'(1/x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}
$$
(if the limits exist, of course), by doing again $x\mapsto 1/x$. Similarly for $x\to-\infty$.
Thus the proofs you need are just two.
Actually, there is a third case (that however includes $\infty/\infty$) and is $\text{(anything)}/\infty$, where (anything) may also be “the limit does not exist”. You find the proof on the Wikipedia page.
A: In trying to understand the other answers, I had to fill in intermediate steps. I want to make those steps explicit here for future reference and to make sure I actually comprehended correctly.
Assume we are able to prove the following result
Result 0

Assume that
$$\lim\limits_{x\to a^+} f(x)=\lim\limits_{x\to a^+} g(x)=0$$
and
$$\lim\limits_{x\to a^+} \frac{f'(x)}{g'(x)}=l$$
Then we can conclude that
$$\lim\limits_{x\to a^+} \frac{f(x)}{g(x)}=\lim\limits_{x\to a^+}
 \frac{f'(x)}{g'(x)}=l$$

(See Spivak, Ch. 11, Theorem 9 for an exact proof of this specific result).
According to the answer from @egreg, we can prove some further similar results using manipulations of limits. Here is an example of one such proof. I had to do all the steps to wrap my head around the operations with limits.
Result 1

Assume that
$$\lim\limits_{x\to a^-} f(x)=\lim\limits_{x\to a^-} g(x)=0$$
and
$$\lim\limits_{x\to a^-} \frac{f'(x)}{g'(x)}=l$$
Then we can conclude that
$$\lim\limits_{x\to a^-} \frac{f(x)}{g(x)}=\lim\limits_{x\to a^-}
 \frac{f'(x)}{g'(x)}=l$$

Let $f_1(x)=f(-x)$ and $g_1(x)=g(-x)$.
Then (I believe according to this result)
$$\lim\limits_{x\to a^+} f_1(x)=\lim\limits_{x\to a^+} f(-x)=\lim\limits_{x\to a^-} f(x)=0$$
$$\lim\limits_{x\to a^+} g_1(x)=\lim\limits_{x\to a^+} g(-x)=\lim\limits_{x\to a^-} g(x)=0$$
Also,
$$\frac{f_1'(x)}{g_1'(x)}=\frac{f'(-x)}{g'(-x)}$$
$$\lim\limits_{x\to a^+}\frac{f_1'(x)}{g_1'(x)}=\lim\limits_{x\to a^+}\frac{f'(-x)}{g'(-x)}=\lim\limits_{x\to a^-}\frac{f'(x)}{g'(x)}$$
According to Result 0 applied to $f_1$ and $g_1$ we have
$$\lim\limits_{x\to a^+}\frac{f_1(x)}{g_1(x)}=\lim\limits_{x\to a^+}\frac{f_1'(x)}{g_1'(x)}=\lim\limits_{x\to a^-}\frac{f'(x)}{g'(x)}\tag{1}$$
But,
$$\lim\limits_{x\to a^+}\frac{f_1(x)}{g_1(x)}=\lim\limits_{x\to a^+}\frac{f(-x)}{g(-x)}=\lim\limits_{x\to a^-}\frac{f(x)}{g(x)}\tag{2}$$
Therefore, putting $(1)$ and $(2)$ together we have
$$\lim\limits_{x\to a^-}\frac{f(x)}{g(x)}=\lim\limits_{x\to a^-}\frac{f'(x)}{g'(x)}$$
