Proof of the L'Hôpital Rule for $\frac{\infty}{\infty}$

I ask for the proof of the L'Hôpital rule for the indeterminate form $\frac{\infty}{\infty}$ utilizing the rule for the form $\frac{0}{0}$.

Theorem: Let $f,g:(a,b)\to \mathbb{R}$ be two differentiable functions such as that: $\forall x\in(a,b)\ \ g(x)\neq 0\text{ and }g^{\prime}(x)\neq 0$ and $\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=+\infty$ If the limit $$\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ exists and is finite, then $$\lim_{x\to a^+}\frac{f(x)}{g(x)}=\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$

My attempt: Since $\lim_{x\to a^+}f(x)=+\infty$, $$\exists \delta>0:a<x<a+\delta<b\Rightarrow f(x)>0\Rightarrow f(x)\neq 0$$ Let $F,G:(a,a+\delta)\to \mathbb{R}$, $F(x)=\frac{1}{f(x)}$, $G(x)=\frac{1}{g(x)}$. Then by the hypothesis $\lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=0$, $$\forall x\in(a,b)\ \ G(x)\neq 0\text{ and }G^{\prime}(x)=-\frac{1}{g^2(x)}g^{\prime}(x)\neq 0$$ The question is, does the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}=\lim_{x\to a^+}\frac{-\frac{1}{f^2(x)}f^{\prime}(x)}{-\frac{1}{g^2(x)}g^{\prime}(X)}=\lim_{x\to a^+}\frac{g^2(x)f^{\prime}(x)}{f^2(x)g^{\prime}(x)}$$ exist?

The limit $$\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ exists by the hypothesis but we don't know if the limit $\displaystyle\lim_{x\to a^+}\frac{g^2(x)}{f^2(x)}$ exists to deduce that the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}$$ exists to use the L'Hôpital Rule for the form $\frac{0}{0}$.

EDIT: After discussing it with other users in the site, we came to the conclusion that this proof is only partial and can't logically be continued to yield the Theorem. As a result, the rule for the $\frac{0}{0}$ form can't be used to proove the rule for the $\frac{\infty}{\infty}$ form. Mr. Tavares and myself have already given two different proofs (with the pretty much the same main idea) of the Theorem in question using Cauchy's Mean Value Theorem. You can read them below. You can also read the proof Rudin gives for a stronger version of the Theorem (that does not suppose that $\lim_{x\to a^+}f(x)=+\infty$) in his book Principle of Meathematical Analysis. If you have any objections in either proofs please let me know. Thank you.

-
This form of the rule does not require that the numerator $\to \infty,$ only the denominator, e.g. see here. –  Bill Dubuque May 20 '12 at 14:43
That just makes it harder to prove... Anyway, I don't care about Rudin's stronger version, you can use the $f\to \infty$ as $x\to a^+$ without second thoughts. –  SomeoneContinuous May 20 '12 at 14:47
Actually the proof isn't much harder, and the extra flexibility proves quite handy in practice. I happen to know a lot about such matters, having studied such topics in depth when I was a student (when I invented the first known algorithm for computing limits and asymptotics for a wide class of functions). –  Bill Dubuque May 20 '12 at 15:59
Can this somehow work in my proof? –  SomeoneContinuous May 20 '12 at 16:22
I will examine Rudin's proof thoroughly at a latter time. For now, how can this specific result be proven? –  SomeoneContinuous May 20 '12 at 16:33

Let $$$$\lim_{x\rightarrow a^{+}}\frac{f^{\prime }(x)}{g^{\prime }(x)}=L. \tag{1}$$$$ Then for each $\delta >0$ there exists a real $\beta \in \left( a,b\right)$ such that for all $x\in \left( a,\beta \right)$ $$$$\left\vert \frac{f^{\prime }(x)}{g^{\prime }(x)}-L\right\vert <\delta . \tag{2}$$$$ Let $x\in \left( a,\beta \right) ,y\in \left( a,\beta \right) ,x<y$. Since the functions $f,g$ are continuous and differentiable on $\left[ x,y\right]$ we can apply the Cauchy Mean Value Theorem. Consequently, there exists a $c\in \left[ x,y\right] \subset \left( a,\beta \right)$ such that $$$$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f^{\prime }(c)}{g^{\prime }(c)}. \tag{3}$$$$ Hence for $x\in \left( a,\beta \right) ,y\in \left( a,\beta \right) ,x<y$ $$\begin{eqnarray} \left\vert \frac{f(x)-f(y)}{g(x)-g(y)}-L\right\vert &<&\delta \\ \\&&\\ \left\vert \frac{f(x)/g(x)-f(y)/g(x)}{1-g(y)/g(x)}-L\right\vert &<&\delta . \tag{4} \end{eqnarray}$$ Assume $\lim_{x\rightarrow a^{+}}f(x)=\lim_{x\rightarrow a^{+}}g(x)=+\infty$ and fix $y$. Then $g(y)/g(x)\rightarrow 0$ and there exists a $\gamma \in \left( a,\beta \right)$ such that for $x\in \left( a,\gamma \right)$, we have $g(x)>0$ and $g(x)/g(y)>1$. Inequality $(4)$ implies $$$$\left( 1-\frac{g(y)}{g(x)}\right) \left( L-\delta \right) <\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}<\left( 1-\frac{g(y)}{g(x)}\right) \left( L+\delta \right) . \tag{5}$$$$ Letting $x\rightarrow a^{+}$ we conclude that $$$$\lim_{x\rightarrow a^{+}}\frac{f(x)}{g(x)}=L. \tag{6}$$$$

Adapted from J. Campos Ferreira, Introdução à Análise Matemática, Teorema 11, p. 386 and J. Santos Guerreiro, Curso de Análise Matemática, Proposição 5.2.3.2, p. 314.

-
I seem to have missed where you used the 0/0 L'Hopital rule anywhere? Can you point that out to me? –  mixedmath May 20 '12 at 17:14
@mixedmath You are right! I've used only the conditions $\lim_{x\rightarrow a^{+}}f(x)=\lim_{x\rightarrow a^{+}}g(x)=+\infty$. –  Américo Tavares May 20 '12 at 17:17
"Since f(x),g(x) are continuous and differentiable functions on (a,x0) there exists c∈(a,x0) such that.." The Cauchy Mean Value Theorem requires that $f,g$ are continuous on a closed interval –  SomeoneContinuous May 20 '12 at 17:18
That is the Cauchy Mean Value Theorem, or am I blind? –  SomeoneContinuous May 20 '12 at 17:22
@Someone: Yes, the Cauchy MVT. –  mixedmath May 20 '12 at 17:24

You're very close to a partial proof. I don't know that the general $\infty / \infty$ rule can be proven from the $0/0$ rule, and it all boils down to exactly the question you ask: does $\lim \dfrac{g^2 f'}{f^2 g'}$ exist? Not always, apriori.

But if we assume it exists, then we know that $\lim \dfrac{g}{f} = \lim \dfrac{1/f}{1/g} = \lim \dfrac{F'}{G'} =\lim \dfrac{F}{G}= \lim \dfrac{g^2 f'}{f^2 g'}$, so that (as we are assuming $\lim f'/g'$ exists and is finite) we may 'cross-multiply' to get that $\lim \dfrac{f}{g} = \lim \dfrac{f'}{g'}$

And so we have a case of the general theorem. (Conceivably, some annoying details may be needed to cover cases where we inadvertently divided by $0$ or whatnot). I do not see how we can use the $0/0$ case to get the complete result. But Américo's answer gives a complete proof independent of the $0/0$ case.

-
I see, it is just that many sources such as cs.uleth.ca/~holzmann/notes/lhospital.pdf suggested that approach. Thank you –  SomeoneContinuous May 20 '12 at 17:38
@Someone: I don't understand. In that link, they prove the $0/0$ and $\infty/\infty$ cases separately, and both with the MVT. –  mixedmath May 20 '12 at 17:42
Am I really that blind? It writes: "In the case $f(x)\to \infty$,$g(x)\to \infty$ define $F(x)=\frac{1}{f(x)}$ and $G(x)=\frac{1}{g(x)}$ and apply the following argument to $F,G$ instead of $f,g$" But in that argument they suppose that the limit exists –  SomeoneContinuous May 20 '12 at 18:01
@Someone: Yes, they imply an (incomplete) argument. It's more or less the same as Américo's answer, which hopefully he'll complete and restore. –  mixedmath May 20 '12 at 18:06

I decided to answer my question by prooving the Theorem. Is this proof correct and fully rigorous?

EDIT: It seems while I was writing this answer Mr. Tavares posted his own. Sorry if anyone was confused

Let $\epsilon>0$ and $\lim_{x\to a^+}\frac{f^{\prime}(x)}{g^{\prime}(x)}=\ell$. Then, $$\exists \delta_0>0:\forall x\in(a,b)\ \ a<x<a+\delta_0\Rightarrow \left|\frac{f^{\prime}(x)}{g^{\prime}(x)}-\ell\right|<\frac{\epsilon}{2}\Rightarrow \frac{f^{\prime}(x)}{g^{\prime}(x)}<\ell+\frac{\epsilon}{2}<\left|\ell\right|+\epsilon\ \ (1)$$ Let $c\in (a,a+\delta_0)$ and $x\in (a,c)$. By Cauchy's Mean Value Theorem for $f|_{[x,c]}$ and $g|_{[x,c]}$, $$\exists \xi_x \in (x,c):\frac{f^{\prime}(\xi_x)}{g^{\prime}(\xi_x)}=\frac{f(c)-f(x)}{g(c)-g(x)}=\frac{f(x)-f(c)}{g(x)-g(c)}\ \ (2)$$ Since $\lim_{x\to a^+}f(x)=+\infty$, $$\exists \delta_1>0:a<x<a+\delta_1<b\Rightarrow f(x)>0\Rightarrow f(x)\neq 0\ \ (3)$$ Let $\delta_2=\min \left\{\delta_0,\delta_1\right\}>0$. Then, for $a<x<a+\delta_2$, (1),(2),(3) hold and so $$\frac{f(x)}{g(x)}=\frac{f(x)}{f(x)-f(c)}\frac{g(x)-g(c)}{g(x)}\frac{f(x)-f(c)}{g(x)-g(c)}\overset{(3)}{=}\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}\frac{f^{\prime}(\xi_x)}{g^{\prime}(\xi_x)}\ \ (4)$$ Since $$\lim_{x\to a^+}\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}=\frac{1-\lim_{x\to a^+}\frac{g(c)}{g(x)}}{1-\lim_{x\to a^+}\frac{f(c)}{f(x)}}=\frac{1-0}{1-0}=1$$ we have that $$\exists \delta_3>0:a<x<a+\delta_3\Rightarrow \left|\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}-1\right|<\frac{\epsilon}{2(\left|\ell\right|+\epsilon)}\ \ (5)$$ Let $\delta=\min \left\{\delta_2,\delta_3\right\}>0$. Then, for $a<x<a+\delta$ (1),(2),(3),(4),(5) hold and so $$\left|\frac{f(x)}{g(x)}-\ell\right|=\left|\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}\frac{f^{\prime}(\xi_x)}{g^{\prime}(\xi_x)}-\ell\right|=\left|\frac{f^{\prime}(\xi_x)}{g^{\prime}(\xi_x)}-\ell+\frac{f^{\prime}(\xi_x)}{g^{\prime}(\xi_x)}\left(\frac{1-\frac{g(c)}{g(x)}}{1-\frac{f(c)}{f(x)}}-1\right)\right|<\frac{\epsilon}{2}+(\left|\ell\right|+\epsilon) \frac{\epsilon}{2(\left|\ell\right|+\epsilon)}=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ Since the choice of $\epsilon$ is arbitary, $$\lim_{x\to a^+}\frac{f(x)}{g(x)}=\ell$$

-