Little-o notation - Knowlpedia

For functions $f$ and $g>0$ on a punctured right neighborhood of zero, $f(\varepsilon)=o(g(\varepsilon))$ as $\varepsilon\downarrow0$ means the following limit vanishes:

\lim_{\varepsilon\downarrow0}\frac{f(\varepsilon)}{g(\varepsilon)}=0.

Equivalently, for every $\eta>0$ there is $\delta>0$ such that $|f(\varepsilon)|\le\eta g(\varepsilon)$ for $0<\varepsilon<\delta$ . A norm replaces absolute value for vector-valued quantities.

Comparison with a bounded ratio

Little-o implies Big-O, but not conversely: $\varepsilon^2=o(\varepsilon)$ , while $\varepsilon$ is not $o(\varepsilon)$ . A uniform little-o statement requires one $\delta$ for all values of every parameter declared uniform.

References

NIST Digital Library of Mathematical Functions, §2.1: Definitions and elementary properties.