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I am looking for a rigorous proof of the following statement:

Fix $f,g : \mathbb{R}\to\mathbb{R}$ with $\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} g(x) = \infty$. Assume that $f', g'$ are differentiable in the right neighbourhood of $0$ and $g'(x)\neq 0$ everywhere sufficiently close to $0$.

Then, if $$\lim_{x\to 0^+} \frac{f'(x)}{g'(x)} = \infty\ ,$$ we also have $$\lim_{x\to 0^+} \frac{f(x)}{g(x)} = \infty\ .$$

All proofs I was able to find seem to either ignore that case or contain mistakes or omissions. One possible route would be to apply the ∞/∞ case of L'Hôpital for a finite limit to $g(x)/f(x)$ (which tends to $0$), but in order to show that $f(x)/g(x)$, one would need something stronger, nemaly that $g(x)/f(x)$ tends to $0^+$, which is not something any version of L'Hôpital's rule that I know of provides.

In fact, what I would ideally like to have is a more ‘unified’ proof of the ∞/∞ case without all these case distinctions (for the 0/0 case, that is possible and quite easy), but I don't think such a proof exists.

UPDATE: Okay, I must have had a momentary blackout. If both $f$ and $g$ tend to infinity, then of course $g(x)/f(x)$ must be positive in a righ neighbourhood of $0$. However, two interesting questions remain:

  1. Does the statement still hold if only one of $f$ and $g$ tends to infinity? For the case where the limit of $f'(x)/g'(x)$ is finite, only $g$ needs to tend to infinity, as shown e.g. in Walter Rudin's proof of L'Hôpital in "Principles of Mathematical Analysis".

  2. Is there some alternative ‘unifying’ approach that avoids all these case distinctions altogether?

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  • $\begingroup$ Well the usual formulation of the de l'Hospital is that the equality holds if the limes of the quotient of the derivatives exist (being either finite or infinite). Given that, in any case one might try to reduce the case $\infty/\infty$ to the second derivatives, etc. $\endgroup$ – Rudi_Birnbaum Aug 17 '16 at 9:03
  • $\begingroup$ if its either one of them then the limit of the quotient also exists, its then either 0 or $\pm\infty$. $\endgroup$ – Rudi_Birnbaum Aug 17 '16 at 9:07
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When $g(x) \rightarrow \infty$ as ${x\rightarrow 0^+}$ then for any $a>0$: $$ I = \liminf_{x\rightarrow 0^+} \frac{f(x)}{g(x)} = \liminf_{x\rightarrow 0^+} \frac{f(x)-f(a)}{g(x)-g(a)} \geq \inf_{0<t<a} \frac{f'(t)}{g'(t)}$$ and the latter was assumed to go to $+\infty$ as $a$ goes to zero.

To see the last inequality let $\lambda=\frac{f(x)-f(a)}{g(x)-g(a)}$. The function $F(t)=f(t)-\lambda g(t)$ verifies $F(t)=F(a)$. By the MVT there is $t\in (x,a)$ for which $F'(t)=f'(t)-\lambda g'(t)=0$. So $f'(t)/g'(t)=\lambda$ which implies our claim

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