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

1. May 20, 2012

### 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 $$\frac{0}{0}$$

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