IMC 1995 · Day 1 · P3

Since $f'$ tends to $-\infty$ and $f''$ tends to $+\infty$ as $x$ tends to $0+$, there exists an interval $(0,r)$ such that $f'(x) < 0$ and $f''(x) > 0$ for all $x \in (0,r)$. Hence $f$ is decreasing and $f'$ is increasing on $(0,r)$. By the mean value theorem for every $0 < x < x_0 < r$ we obtain $$f(x) - f(x_0) = f'(\xi)(x - x_0) > 0,$$ for some $\xi \in (x, x_0)$. Taking into account that $f'$ is increasing, $f'(x) < f'(\xi) < 0$, we get $$x - x_0 < \frac{f'(\xi)}{f'(x)} (x - x_0) = \frac{f(x) - f(x_0)}{f'(x)} < 0.$$ Taking limits as $x$ tends to $0+$ we obtain $$-x_0 \le \liminf_{x \to 0+} \frac{f(x)}{f'(x)} \le \limsup_{x \to 0+} \frac{f(x)}{f'(x)} \le 0.$$ Since this happens for all $x_0 \in (0,r)$ we deduce that $\lim\limits_{x \to 0+} \dfrac{f(x)}{f'(x)}$ exists and $$\lim_{x \to 0+} \frac{f(x)}{f'(x)} = 0.$$