How is this de l'Hospital's rule simplified First I proved the de l'Hospital's rule here. The rule is:
If $f,g : (a,b) \to \mathbb R$ are differentiable and $\lim_{x \to a^+}f(x) = \lim_{x \to a^+} g(x) = 0 $ and $\lim_{x \to a^+} {f'(x) \over g'(x)} = L$ then $\lim_{x \to a^+} {f(x) \over g(x)} = L$.
Then I proved a simplified de l'Hospital's rule here. The rule is:
If $f,g: [a,b) \to \mathbb R$ are differentiable and $\lim_{x \to a^+}f(x) = \lim_{x \to a^+} g(x) = 0 $ and $\lim_{x \to a^+} {f'(x) \over g'(x)} = L$. Also assume that $f'$ and $g'$ are continuous at $a$. Then $\lim_{x \to a^+} {f(x) \over g(x)} = L$.
The problem is: the proof of the simplified rule is not simplified. It is the same as the proof for the not simplified rule. Where is my mistake?
 A: In case you know the limit of the derivatives (for $x\rightarrow a$) exists, and that $g^\prime(a)\neq 0$ (are you assuming that?), you can argue much (?) easier as follows:
$$\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}= \frac{f(x)-f(a)}{x-a} \frac{x-a}{g(x)-g(a)}$$
and now you can use the fact that the product is continuous, so if the limits exist the product will converge to the expected result.
This is in fact much easier than your proof, since you don't need to know the general Cauchy version of the mean value theorem (the one with general $g(x)$ different from just $x$ -- you may recall that a rather experienced users comment showed that he was not aware of that version). And of course this reasoning fails if $g^\prime(a) =0$, which is irrelevant for the other proof.
(I doubt you can do better in case $g^\prime(a) = 0$, then you will need the more 'advanced' reasoning. If you know it anyway, why do you care?)
A: I believe you have given the statement of l'Hospital's rule in the first statement and your are taking the converse of it in the second case.  
But the proof doesnt allow you to go that way. I have just given the proof in a brief way, in this proof you cannot go in the backward direction, so your second statement is false.  
Take two functions $f(x)$ and $g(x)$ defined over $[a,a+h]$ and $f(a)=g(a)=0$. Now as per the cauchy mean  value theorem there exists a $c$ that belongs to $(a,a+h)$ such that it satisfies
$$\dfrac{f'(c)}{g'(c)} = \dfrac{f(a+h)-f(a)}{g(a+h)-g(a)}.$$
As $f(a)=g(a)=0$ It gets simplified to
$$\dfrac{f'(c)}{g'(c)} = \dfrac{f (a+h)}{g(a+h)}.$$
Now apply the limit as $h$ tends to zero you get the l'Hospital's rule.
