 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/15/c/ii/solution

The definition of hyperbolic sine simplifies the composition exactly for

x > 0

sinh (ln x) = \frac{1}{2} (e^{l n x} - e^{- l n x}) = \frac{1}{2} (x - x^{- 1}) .

h = x - 1

1/ (1 + h) = 1 - h + h^{2} - h^{3} + h^{4} + O (h^{5})

, valid for

∣ h ∣ < 1

, then gives

sinh (ln x) = (x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{2} (x - 1)^{3} - \frac{1}{2} (x - 1)^{4} + O ((x - 1)^{5}) .

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