Past exam of the differential equations 2 course of the University of Oxford / 2015 / 3 / a / Solution

By the definition of little o notation, the required quotient isAs $ϵ\to 0^{+}$, write this as $\log ϵ/(1/ϵ)$. L'Hôpital's rule applies to this infinite-over-infinite ratio and givesTherefore ${ϵ}^2\log ϵ=o(ϵ)$. 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.^[ref]