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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-4/3e/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 3E / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

f_{n} (x) = x^{2} + \frac{1}{n} .

(1)

Each

f_{n}

is a differentiable function on

[- 1, 1]

, while

0 \leq f_{n} (x) - ∣ x ∣ = \frac{1/ n}{x ^{2} + 1/ n + ∣ x ∣} \leq \frac{1}{n} .

(2)

Thus $f_{n}$ converges uniformly to the absolute value function $∣ x ∣$ , which is not differentiable at $0$ . Uniform convergence alone therefore does not preserve differentiability.

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