Let f:(0,∞)→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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