Show that $\lim_{x\to \infty} \frac{f(cx)}{f(x)}=1, ∀c>0$.

Question

Lef $f\colon \mathbb{R}\to \mathbb{R}$ a monotone function such that $\lim_{x \to \infty} \frac{f(2x)}{f(x)} =1$. Show that $\lim_{x \to \infty} \frac{f(cx)}{f(x)}=1, ∀c>0$.

Answer 1

First show this for $c$ a power of $2$, by breaking it into a product of ratios like $\frac{f(8x)}{f(x)} = \frac{f(2\cdot 4x)}{f(4x)}\cdot\frac{f(2\cdot 2x)}{f(2x)}\cdot\frac{f(2\cdot x)}{f(x)}$. The use monotonicity to finish it off for arbutrary $c$ by choosing a power of $2$, say $2^n$, that is greater than $c$ so that $f(cx)$ is between $f(x)$ and $f(2^nx)$.