TITLE: $f:\mathbb{R} \mapsto \mathbb{R}$ convex and increasing. Show there is $a \in \mathbb{R}$ such that $\lim\limits_{x\to -\infty}\frac{f(x)}{x}=a$
QUESTION [5 upvotes]: Let $f:\mathbb{R} \to \mathbb{R}$ be a convex and increasing function such that $\lim\limits_{x\to -\infty}f(x)\to-\infty$ Show that there is an $a \in \mathbb{R}$ such that $\lim\limits_{x\to -\infty}\frac{f(x)}{x}=a$.
First of all, i can't really find an example respecting the hypothesis. Is there a convex function which is increasing with $\lim\limits_{x\to -\infty}f(x)=-\infty$? If $f$ was supposed concave, it appears true to me.
Then, if the statement holds, I don't really see how to prove it. I tried to apply the following property of convex function on the interval $x<a<b$: $$\frac{f(a)-f(x)}{a-x}\le\frac{f(b)-f(x)}{b-x}\le\frac{f(b)-f(a)}{b-a}.$$
But, even if i suceed to "sandwich" ${f(x) \over x}$, i don't really see how it could help.. Thank you in advance for help.

REPLY [3 votes]: We know that $f(c) \le 0$ for some $c \in \Bbb R$. For $x < y < c$ we have $f(x)  \le 0$ and the convexity condition
$$
 f(y) \le \frac{c-y}{c-x} f(x) + \frac{y-x}{c-x} f(c) \le \frac{c-y}{c-x} f(x) \, .
$$
It follows that
$$
 0 \le \frac{f(x)}{x-c} \le \frac{f(y)}{y-c} 
$$
which shows that $f(x)/(x-c)$ decreases with decreasing $x$ and is bounded below by $0$. Therefore
$$
 a = \lim_{x \to -\infty} \frac{f(x)}{x-c}
$$
exists, and then
$$
 \frac{f(x)}{x} = \frac{x-c}{x} \frac{f(x)}{x-c}
$$
has the same limit $a$ for $x \to -\infty$.