 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/7e/a/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

x_{n} = 12 - \frac{1000 n ^{2} + 4}{1. 1 ^{n}} .

For

n \geq 3

1. 1^{n} = (1 + 0.1)^{n} \geq (3 n) (0.1)^{3} = \frac{n ( n - 1 ) ( n - 2 )}{6000} .

Consequently

(1000 n^{2} + 4) /1. 1^{n} \to 0

: given

ε > 0

, the displayed rational upper bound is below

ε

for all sufficiently large

n

. Thus, directly from the definition of convergence,

x_{n} ⟶ 12 .

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