$\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 $

Question

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.

Answer 1

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 $\lim_{x \to a^+} f(x) = +\infty$.

Answer 2

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.