If $\lim_{x\to 0^+}f'(x)=-\infty$ and $\lim_{x\to 0^+}f''(x)=\infty$ it is always the case that $\lim_{x\to 0^+}\frac{f(x)}{f'(x)}=0$? My teacher proposed the following question in our class:

Let $f$ be a twice differentiable function in $\mathbb{R}^+$ such that
  $\lim_{x\to 0^+}f'(x)=-\infty$ and $\lim_{x\to 0^+}f''(x)=\infty$. Find the value of $$\lim_{x\to 0^+}\frac{f(x)}{f'(x)}$$ or show that it doesn't exist.

I found that both $f(x)=-\ln(x)$ and $f(x)=\frac{1}x$ satisfy the conditions and make the proposed limit equal to $0$. How do I prove that this is always the case? Thanks.

Let $x_0>0$ with $f'<0,f''>0 $ on $(0,x_0).$ We have 
$$\tag 1\left | \frac{f(x)}{f'(x)}\right| \le \left | \frac{f(x)-f(x_0)}{f'(x)}\right | + \left |\frac{f(x_0)}{f'(x)}\right|.$$
The second summand on the right tends to $0$ as $x\to 0^+.$  It follows that
$$\tag 2 \limsup_{x\to 0^+}\left |\frac{f(x)}{f'(x)}\right| \le \limsup_{x\to 0^+}\left | \frac{f(x)-f(x_0)}{f'(x)}\right |.$$
By the mean value theorem, the fraction on the right side of $(2)$ equals
$$\tag 3 \frac{|f'(c_x)||x-x_0|}{|f'(x)|}.$$
Now $f''>0$ on $(0,x_0)$ implies $f'$ increases there. Thus $(3)$ is bounded above by
$$\le \frac{|f'(x)||x-x_0|}{|f'(x)|} = |x-x_0| < x_0.$$
It follows that the left side of $(2)$ is $\le x_0.$ Since $x_0$ is arbitrarily small, the the left side of $(2)$ is $0.$ This proves $f(x)/f'(x) \to 0$ as desired.
NOTE (added later): I did not use $f''\to \infty,$ only that $f''>0$ near $0.$
