# $f\geqslant 0$, $f''\leqslant0$ and $f$ is differentiable on $]1,\infty[$. $\lim_{x\to\infty} \frac{f(x)}{x}$ exist.

Let $$f\geqslant 0$$, $$f''\leqslant0$$ and $$f$$ is differentiable on $$]1,\infty[$$. Show that $$\lim_{x\to\infty} \frac{f(x)}{x}$$ exist.

I already asked quite similar question, but this time i'd like to know how to proceed on interval $$]1,\infty[$$. I was trying to consider $$1 and apply Mean value theorem on intervals $$]a,x[$$ and $$]x,b[$$ and then try to ''play'' with inequalities as we know that $$f'$$ is decreasing. But, i don't really know if it could lead me somewehere.. Thanks in advance

We could probably resolve this problem with Bernouilli-Hospital, but i'm interested in Mean value proof if it is possible.

• You can use the same methods as in your previous question math.stackexchange.com/q/3976654 to show that $\lim_{x\to\infty} \frac{f(x)}{x-2}$ exists. Then conclude that $\lim_{x\to\infty} \frac{f(x)}{x}$ exists as well. Jan 18, 2021 at 13:20
• Well, that's the point: $f'$ is weakly decreasing therefore $\lim_{x\to\infty}f'(x)=c\in\Bbb R\cup\{-\infty\}$. Since $f\ge0$, $f'$ must be $\ge0$, because if $f'(\alpha)<0$ then $f(x)\le f(\alpha)+f'(\alpha)(x-\alpha)\to -\infty$ for $x\ge\alpha$. Therefore $f$ is weakly increasing. If $f$ is bounded, then $\frac{f(x)}x\to 0$. If $f$ is unbounded, then by being weakly increasing we have $f\to\infty$ and you can use L'Hopital $\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}f'(x)= c$. Or you can work a bit with Lagrange until you prove the result that L'Hopital gives in a snap of fingers.
– user239203
Jan 18, 2021 at 13:22

For $$1 < a < x < y$$ there are $$c_1 \in(a, c)$$ and $$c_2 \in (x, y)$$ such that $$\frac{f(y)-f(x)}{y-x} = f'(c_2) \le f'(c_1) = \frac{f(x)-f(a)}{x-a} \le \frac{f(x)}{x-a} \, .$$ This inequality can be rearranged to $$\frac{f(x)}{x-a} \ge \frac{f(y)}{y-a}$$ which shows that $$\frac{f(x)}{x-a}$$ is decreasing. It is also bounded below by zero, so that $$\lim_{x \to \infty} \frac{f(x)}{x-a}$$ exists. But then $$\frac{f(x)}{x} = \frac{x-a}{x} \cdot \frac{f(x)}{x-a}$$ also has a limit for $$x \to \infty$$.