Upper bound of natural logarithm for expression close to 1

combinatoricsinequalitylogarithmsnumber theoryreal-analysis

Say $0 \leq a,b \leq 1$ and $a$ and $b$ are $\epsilon$-close. Specifically, $|a-b| \leq \epsilon$ for some small $\epsilon > 0$. Consider

$$
X = \lvert \ln \frac{a}{b} \rvert.
$$

In what circumstances can we upper bound this expression in terms of $\epsilon$? What are the upper bounds in those cases?

Thoughts: it's clear that if $\epsilon \to 0$, then $\frac{a}{b}\to1$ and $X\to0$ but not sure how to move forward.

Best Answer

If $a$ is close to $0$ there is no hope so assume that $a \geq c$ for some $c>0$. Then $\log(\frac a b)|=|\log\,a -\log \, b|=|a-b|\frac 1 t$ (by MVT) for some $t \in (a,b)$. Hence $|\log(\frac a b)|<\epsilon \frac 1 c$.

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