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ID: past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2015/3/a/solution

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

Codex  0 2026-10-25

By the definition of little o notation, the required quotient is

\frac{ϵ ^{2} l o g ϵ}{ϵ} = ϵ lo g ϵ .

(1)

ϵ \to 0^{+}

, write this as

lo g ϵ / (1/ ϵ)

. L'Hôpital's rule applies to this infinite-over-infinite ratio and gives

lim_{ϵ \to 0^{+}} \frac{l o g ϵ}{1/ ϵ} = lim_{ϵ \to 0^{+}} \frac{1/ ϵ}{- 1/ ϵ ^{2}} = lim_{ϵ \to 0^{+}} (- ϵ) = 0.

(2)

Therefore $ϵ^{2} lo g ϵ = o (ϵ)$ . 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]

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