If $f,f':(0,\infty) \rightarrow (0,\infty)$, $f, f' \rightarrow 0$ as $x \rightarrow 0^{+}$, does $f/f' \rightarrow 0$?

Question

To be explicit: $f:\mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function, $f(x),f'(x) > 0$ for $x > 0$, and $$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} f'(x) = 0$$ Can we conclude that: $$\lim_{x \rightarrow 0^{+}} \frac{f(x)}{f'(x)} = 0$$ If we assume $f$ is $C^{k}$ in a neighborhood of $0$ and $f^{(k)}(0) \neq 0$, then we can prove it using induction and l'Hospital. Perhaps without these assumptions there's some sort of oscillating sine-wave-esque counterexample? Thoughts?

Answer 1

You're right, a sine-wave-esque thing will be a counterexample. Sketch: Define

$$g(t) = t\sin^2(\pi/t) + t^3,\,\, f(x) = \int_0^x g(t)\,dt.$$

Then $f$ and $f'=g$ satisy the hypotheses. But note

$$f(1/n) > \sum_{k=n}^{\infty}\int_{1/(k+1)}^{1/k} t\sin^2(\pi/t)\,dt.$$

Exercise: Verify the last sum is on the order of $1/n^2$ as $n\to \infty.$ On the other hand, $f'(1/n)=g(1/n) = 1/n^3.$ Thus $f/f'(1/n)$ is on the order of $n \to \infty.$