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Let $a \in \mathbb R $ and $f,g:]a$,+∞[$\to \mathbb R$ two differentiable functions such that $g'$ is never equal to $0$. I need to prove that $\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = +\infty \implies \lim_{x \to a^+}\frac{f'(x)}{g'(x)} \neq -\infty $

I tried to use mean value theorem but it does not lead me anywhere. I am also not very familiar with proving things with $\neq$...

I am not allowed to use L'Hopital.

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2 Answers 2

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According to Darboux's theorem, as $g^\prime$ never vanishes it is always positive or always negative. $\lim_{x \to a^+} g(x) = +\infty$ implies that $g^\prime(x) \lt 0$ for $x \in (a, \infty)$.

$\lim_{x \to a^+}\frac{f^\prime(x)}{g^\prime(x)} = -\infty $ would imply the existence of $\delta \gt 0$ such that $f^\prime(x) \gt 0$ for $x \in (a, a+ \delta)$. A contradiction with $\lim_{x \to a^+} f(x) = +\infty$.

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    $\begingroup$ Suppose that $f = \frac1{x}$, $a=0$. Hence $f' < 0$ for $x \in (0,1)$ and $f(x) \to \infty$, $x \to 0+$. So it doesn't look like contradiction. $\endgroup$ Jan 5, 2021 at 19:36
  • $\begingroup$ @BotnakovN. Thanks for pointing the sign errors. Fixed now. $\endgroup$ Jan 5, 2021 at 19:49
  • $\begingroup$ Now everything is OK. I also added "+" in front of "$\infty$" in two places, because formally $\lim = \infty$ may be in case $\lim =- \infty$. $\endgroup$ Jan 5, 2021 at 19:55
  • $\begingroup$ @BotnakovN. Ok with this despite I thought that $\infty$ without a sign meant $+\infty $ in English. $\endgroup$ Jan 5, 2021 at 20:04
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Suppose $f'(x)/g'(x) \to -\infty.$ Then there exists $b>a$ such that $f'(x)/g'(x) <0$ for $x\in (a,b).$ But for $x$ close to $a,$

$$\frac{f(x)-f(b)}{g(x)-g(b)}=\frac{f'(c_x)}{g'(c_x)}$$ by Cauchy's MVT. Note the left side is positive for such $x$ (because $f,g\to +\infty$) while the right side is negative. This contradiction shows $f'(x)/g'(x) \to -\infty$ cannot happen.

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