$\lim_{x \to a^+} f(x) = \lim_{x \to a^+} g(x) = +\infty \implies \lim_{x \to a^+}\frac{f'(x)}{g'(x)} \neq -\infty$

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.

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 $$\lim_{x \to a^+} f(x) = +\infty$$.
• 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. Jan 5, 2021 at 19:36
• Now everything is OK. I also added "+" in front of "$\infty$" in two places, because formally $\lim = \infty$ may be in case $\lim =- \infty$. Jan 5, 2021 at 19:55
• @BotnakovN. Ok with this despite I thought that $\infty$ without a sign meant $+\infty$ in English. Jan 5, 2021 at 20:04
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.