 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/19/b/solution

sin (1/ x_{n}) = sin (2 nπ + π /6) = 1/2

, whereas along the second sequence,

sin (1/ x_{n}^{'}) = sin (2 nπ) = 0

. The outer sine therefore gives

f_{0} (x_{n}) = sin (π /2) = 1, f_{0} (x_{n}^{'}) = 0 .

Both sequences approach zero through positive values. For arbitrary

a

, the same calculation gives

f_{a} (x_{n}) = x_{n}^{a}

and

f_{a} (x_{n}^{'}) = 0

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