 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-1/10f/b/solution

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

Let

m_{n} = min_{0 \leq x \leq n^{- 1/2}} f (x), M_{n} = max_{0 \leq x \leq n^{- 1/2}} f (x) .

Continuity at zero gives

m_{n}, M_{n} \to f (0)

. Part (a), with the positive weight

n e^{- n x}

, gives

m_{n} (1 - e^{- n}) \leq \int_{0}^{1/ n} n f (x) e^{- n x} d x \leq M_{n} (1 - e^{- n}) .

The squeeze theorem therefore yields

\int_{0}^{1/ n} n f (x) e^{- n x} d x \to f (0) .

Since

f

is bounded on

[0, 1]

, say

∣ f ∣ \leq K

, the omitted tail satisfies

\int_{1/ n}^{1} n f (x) e^{- n x} d x \leq K (e^{- n} - e^{- n}) ⟶ 0.

Adding the tail proves

\int_{0}^{1} n f (x) e^{- n x} d x \to f (0) .

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