Proof of the L'Hôpital Rule for the Indeterminate Form

[3.1415926535]

I ask for the Proof of the L'Hôpital Rule for the Indeterminate Form \frac{\infty}{\infty} utilising the Rule for the form \frac{0}{0}

The 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) 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 $$\lim_{x\to a^+}\frac{g^2(x)}{f^2(x)}$$ exists to deduce that the limit $$\lim_{x\to a^+}\frac{F^{\prime}(x)}{G^{\prime}(x)}$$ exists and use the L'Hôpital Rule for the form \frac{0}{0}

 

[fresh_42]

This is not as easy as that.
https://de.wikibooks.org/wiki/Beweisarchiv:_Analysis:_Differentialrechnung:_L'Hospitalsche_Regel
(It's in the wrong language but the formulas should speak for themselves, otherwise "Google translate in chrome" can be used).

 

[wrobel]

I would propose the following version of this theorem. Let $f,g$ be two continuous in $[0,\infty)$ functions (this condition can be relaxed) such that
$$\lim_{x\to\infty} f(x)=f_0,\quad \lim_{x\to\infty}g(x)=g_0\ne 0.$$
Then
$$\lim_{x\to\infty}\frac{\int_0^xf(s)ds}{\int_0^xg(s)ds}=\frac{f_0}{g_0}.$$
It directly follows from a fact about the mean value:
$$\lim_{x\to\infty} f(x)=f_0\Longrightarrow \lim_{x\to\infty}\frac{1}{x}\int_0^xf(s)ds=f_0$$ which is nice by itself and easy to proof.

Some special cases are remained to think about