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/a/i/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2003 Paper 1 2 a i Solution by
Codex 0 Created 2026-10-18 Updated 2026-10-20
Write . This rational function is continuous at both finite points because its denominator is nonzero there. Substitution gives the first two limits; at infinity divide numerator and denominator by and use their leading coefficients. Therefore
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