Prove $\lim\limits_{x\to 0}f(x)/x=0$ if $\lim \limits_{x\to 0}f(x)=0$ and $\lim\limits_{x\to 0}\frac{f(x)-f(x/2)}{x}=0$, what to do with $<2\epsilon$?

Question

This is problem 3.2.7. from Problems in Real Analysis, Radulescu et. al., I'm stuck on the very last part.

The problem is defined such that $x$ is in the open interval $(0,1)$. From $\lim\limits_{x\to 0} \frac{f(x)-f\left(x/2\right)}x=0$, given an $\epsilon$, we have: $$|f(x)-f\left(\frac{x}{2}\right)|<\varepsilon x$$ from which we can obtain $|f(\frac{x}{2^n})-f(\frac{x}{2^{n+1}})|< \varepsilon\frac{x}{2^n}$ and using the triangle inequality we get $$|f(x)-f\left(\frac{x}{2^n}\right)| \le 2\varepsilon x.$$ If we take $n\to\infty$, since $\lim\limits_{x\to 0} f(x)=0$, we get: $$|f(x)|\le 2\varepsilon |x|$$ $$\frac{|f(x)|}{|x|} \le 2\varepsilon$$

My question is, how come this implies that $\lim \limits_{x \to 0} \frac{f(x)}{x} = 0$? Don't we still need to go from $\frac{|f(x)|}{|x|} \leq 2\varepsilon$ to $\frac{|f(x)|}{|x|} < \varepsilon$?

Answer 1

Note that $\frac{|a|}{|b|} = \left|\frac{a}{b}\right|,$ so the inequality you have and the inequality you want are equivalent.