= Solution
By the definition of <little o notation>, the required quotient is
$$
\frac{\epsilon^2\log\epsilon}{\epsilon}=\epsilon\log\epsilon.
$$
As $\epsilon\to0^+$, write this as $\log\epsilon/(1/\epsilon)$. <L'Hôpital's rule> applies to this infinite-over-infinite ratio and gives
$$
\lim_{\epsilon\to0^+}\frac{\log\epsilon}{1/\epsilon}
=\lim_{\epsilon\to0^+}\frac{1/\epsilon}{-1/\epsilon^2}
=\lim_{\epsilon\to0^+}(-\epsilon)=0.
$$
Therefore \b[$\epsilon^2\log\epsilon=o(\epsilon)$]. The <logarithm> is negative on this approach, which is consistent with a quotient tending to zero from below; <little o notation> concerns the limiting ratio, not its sign.<NIST Digital Library of Mathematical Functions/2.1 Definitions and Elementary Properties>[§2.1(i), equation 2.1.2]{ref}
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