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

  1. May 20, 2012 #1
    I ask for the Proof of the L'Hôpital Rule for the Indeterminate Form [tex]\frac{\infty}{\infty}[/tex] utilising the Rule for the form [tex]\frac{0}{0}[/tex]

    The Theorem: Let [tex]f,g:(a,b)\to \mathbb{R}[/tex] be two differentiable functions such as that:
    [tex]\forall x\in(a,b)\ \ g(x)\neq 0\text{ and }g^{\prime}(x)\neq 0[/tex] and [tex]\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=+\infty[/tex]
    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 [tex]\lim_{x\to a^+}f(x)=+\infty[/tex], $$\exists \delta>0:a<x<a+\delta<b\Rightarrow f(x)>0\Rightarrow f(x)\neq 0$$
    Let [tex]F,G:(a,a+\delta)[/tex] [tex]F(x)=\frac{1}{f(x)}[/tex], [tex]G(x)=\frac{1}{g(x)}[/tex] Then by the hypothesis [tex]\lim_{x\to a^+}F(x)=\lim_{x\to a^+}G(x)=0[/tex] $$\forall x\in(a,b)\ \ G(x)\neq 0\text{ and }G^{\prime}(x)=-\frac{1}{g^2(x)}g^{\prime}(x)\neq 0$$
    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 [tex]\frac{0}{0}[/tex]
  2. jcsd
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