Prove that $\lim\limits_{x\to0^+}\frac{f(x)}{f'(x)}=0$. 
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?

There's c so that when $0\lt x \lt c$, function $f'(x)$ is monotonic increasing.
Rewrite $f(x)=f(c)-\int_x^c f'(x) dx$
For any $\epsilon \gt 0$, there must be $\delta_1, \delta_2$ so that when $0 \lt x\lt \delta_1$, $\left|\frac{f(c)}{f'(x)}\right| \lt \frac{\epsilon}3$ and when $0\lt x\lt \delta_2$, $\left|\frac{f'(\frac{\epsilon}{3})}{f'(x)}\right|<\frac{\epsilon}{3c}$
Let $\delta= \min\{\frac{\epsilon}3,\delta_1,\delta_2\}$, so for $0\lt x\lt \delta$
$\left|\frac{f(x)}{f'(x)}\right|=\left|\frac{f(c)-\int_x^c f'(x)dx}{f'(x)}\right|=\left|\frac{f(c)-\int_x^c f'(x)dx}{f'(x)}\right|=\left|\frac{f(c)-\int_x^{\frac{\epsilon}3} f'(x)dx+\int_{\frac{\epsilon}3}^c f'(x)dx}{f'(x)}\right|\le \left|\frac{f(c)}{f'(x)}\right|+\left|\frac{\int_x^{\frac{\epsilon}3} f'(x)dx}{f'(x)}\right|+\left|\frac{\int_{\frac{\epsilon}3}^c f'(x)dx}{f'(x)}\right|\le \epsilon$
