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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-ii/2/a/solution

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 2 / a / Solution by

Codex  0 2026-10-25

(a_{n})

has limit of a sequence

L

if

\forall ε > 0 \exists N \in N \forall n \geq N : ∣ a_{n} - L ∣ < ε .

For the reciprocal rule, the required condition is

L \neq = 0

. Indeed, eventually

∣ a_{n} - L ∣ < ∣ L ∣/2

, so

∣ a_{n} ∣ \geq ∣ L ∣/2 > 0

. The reciprocals are then defined on this tail, and

\frac{1}{a _{n}} - \frac{1}{L} = \frac{∣ a _{n} - L ∣}{∣ a _{n} ∣∣ L ∣} \leq \frac{2∣ a _{n} - L ∣}{∣ L ∣ ^{2}} ⟶ 0.

A finite number of zero initial terms does not affect this tail limit of a sequence; if a reciprocal sequence is required at every index, those terms must be excluded or assigned arbitrary values separately.

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