 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-1/2/b/ii/solution

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

Both numerator and denominator have a double zero at one. Applying L'Hôpital's rule twice gives

x \to 1 lim \frac{1 - sin ( π x /2 )}{x ^{3} - 3 x + 2} = x \to 1 lim \frac{- ( π /2 ) cos ( π x /2 )}{3 x ^{2} - 3} = x \to 1 lim \frac{( π ^{2} /4 ) sin ( π x /2 )}{6 x} = \frac{π ^{2}}{24} .

Alternatively

x^{3} - 3 x + 2 = (x - 1)^{2} (x + 2)

and

1 - sin (π x /2) \sim π^{2} (x - 1)^{2} /8

, giving the same result from a Taylor expansion.

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