Exercise 1480Calculus › Single-variable calculus › Differentiation › Differentiable functionsL1Added 2 Nov 2026

A convex function and its asymptote

#convexity#slopes#asymptotes

Statement

Let $f:\mathbb{R}\to\mathbb{R}$ be convex. Assume $\dfrac{f(x)-f(0)}x\to\ell\in\mathbb{R}$ as $x\to+\infty$ . Show that $\dfrac{f(x)}x\to\ell$ and that $x\mapsto f(x)-\ell x$ is non-increasing.

Hint

For a convex function, the slope $\dfrac{f(z)-f(y)}{z-y}$ is non-decreasing in $z$ for fixed $y$ . Let $z\to+\infty$ .

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Solution

First claim.
$\frac{f(x)}x=\frac{f(x)-f(0)}x+\frac{f(0)}x\longrightarrow\ell+0=\ell.$

Slopes at infinity. Fix $y\in\mathbb{R}$ . For $z>y$ ,
$\frac{f(z)-f(y)}{z-y}=\frac z{z-y}\cdot\frac{f(z)}z-\frac{f(y)}{z-y}\longrightarrow1\cdot\ell-0=\ell\quad(z\to+\infty).$

Convexity. For fixed $y$ , the slope $z\mapsto\frac{f(z)-f(y)}{z-y}$ is non-decreasing on $(y,+\infty)$ , so every slope is at most its limit $\ell$ . Therefore for $x>y$ ,
$\frac{f(x)-f(y)}{x-y}\le\ell\ \Longrightarrow\ f(x)-\ell x\le f(y)-\ell y.$

Conclusion. $x\mapsto f(x)-\ell x$ is non-increasing.

A convex function and its asymptote

Statement

Solution

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