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.
On
There is no upper bound.
From your bound on $|a - b|$, it follows that
$-\epsilon < a - b < \epsilon$, which implies
$-\epsilon + b < a < \epsilon + b$;
Hence,
$\frac{a}{b}$ might be very close in value to $\frac{\epsilon + b}{b}$ which tends to $\infty$ as $b$ approaches $0$.
Therefore, $\ln(a/b)$ would ``crawl to'' but tend to $\infty$ nevertheless.
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 $|\log(\frac a b)|<\epsilon \frac 1 c$.