real analysis - $f$ is convex, increasing then $lim_{xrightarrow -infty }f(x)/x$ exists

I am stuck with the following question:

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be convex and increasing and $\displaystyle\lim_{x\rightarrow -\infty}f(x)=-\infty$. Prove that there exists $\alpha_{0}\in\mathbb{R}$ such that

$$\lim_{x\rightarrow -\infty}\dfrac{f(x)}{x}=\alpha_{0}.$$

Can someone give me a hint?

Thanks.

Answer

We can show that the limit

$$\lim_{x\to-\infty} \frac{f(0)-f(x)}{0-x}$$
exists.
On the one hand, $\frac{f(0)-f(x)}{0-x} >0$ because $f$ is an increasing function. On the other hand, the convexity implies that $\frac{f(0)-f(x)}{0-x}$ decreases (or at least does not increase) for $x\to -\infty$. A non-increasing function which is bounded from below has a limit. We define

$$a_0 = \lim_{x\to-\infty} \frac{f(0)-f(x)}{0-x}$$
and we have
$$\lim_{x\to-\infty} \frac{f(x)}{x} \\ = \lim_{x\to-\infty} \left(\frac{f(0)-f(x)}{0-x} - \frac{f(0)}{-x}\right) \\ = \lim_{x\to-\infty} \frac{f(0)-f(x)}{0-x} - \lim_{x\to-\infty}\frac{f(0)}{-x} \\ = a_0 - 0 \\ = a_0$$