 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/1/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The lower-order terms cancel, so the numerator must be expanded through sixth order. The Taylor series give

sin (sin x) = x - x^{3} /3 + x^{5} /10 + O (x^{7})

and

sin^{2} x = x^{2} - x^{4} /3 + 2 x^{6} /45 + O (x^{8})

. Therefore

\frac{x s i n ( s i n x ) - s i n ^{2} x}{x ^{4}} = \frac{x ^{2}}{18} + O (x^{4}) .

The first nonzero term is $x^{2} /18$ , with the quotient continuously extended to zero if needed.

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