Let $f:(0,\infty)\to\mathbb{R}$ be a twice differentiable function with $f''$ continuous and let $\lim\limits_{x\to0^+}f'(x)=-\infty$ and $\lim\limits_{x\to0^+}f''(x)=+\infty$. Prove that:
$$\lim_{x\to0^+}\frac{f(x)}{f'(x)}=0.$$
My problem is not a proof of this itself (e.g. using $\epsilon-\delta$ definition). I recently found this in an old high-school textbook where no mention of the "traditional" $\epsilon-\delta$ definition is made, so, is it possible to find a solution without it?
What we can do is find some $a>0$ such that $f$ is strictly decreasing and $f'$ strictly increasing in $(0,a)$ which proves that
$$\lim_{x\to0^+}f(x)=\ell$$
exists (either number or $+\infty$) and we can easily prove what we want in case $\ell\in\mathbb{R}$. But that case $\ell=+\infty$ is one I cannot solve without proving some inequality of the form:
$f(x)+\epsilon f'(x)<0,$
for $x\in(0,\delta)$ for some $\delta>0$. But this is not supposed to be the solution in a high school textbook.
So, does anyone have a more "elementary" solution or an appropriate rephrasing of a current one?
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