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

Solution 1:

Let $$\begin{equation} \lim_{x\rightarrow a^{+}}\frac{f^{\prime }(x)}{g^{\prime }(x)}=L. \tag{1} \end{equation}$$ 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) $ $$\begin{equation} \left\vert \frac{f^{\prime }(x)}{g^{\prime }(x)}-L\right\vert <\delta . \tag{2} \end{equation}$$ 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 $$\begin{equation} \frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f^{\prime }(c)}{g^{\prime }(c)}. \tag{3} \end{equation}$$ 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 $$\begin{equation} \left( 1-\frac{g(y)}{g(x)}\right) \left( L-\delta \right) <\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}<\left( 1-\frac{g(y)}{g(x)}\right) \left( L+\delta \right) . \tag{5} \end{equation}$$ Letting $x\rightarrow a^{+}$ we conclude that $$\begin{equation} \lim_{x\rightarrow a^{+}}\frac{f(x)}{g(x)}=L. \tag{6} \end{equation}$$

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.

Solution 2:

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.

Solution 3:

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, $$\lim_{x\to a^+}\frac{f(x)}{g(x)}=\ell$$