 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2013-resit/1/solution

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

The constant, linear and quadratic terms cancel in the numerator. Consequently

1 + ln (1 + x) - 1 + 2 x = - \frac{x ^{3}}{6} + O (x^{4}), x \to 0 lim \frac{1 + ln ( 1 + x ) - 1 + 2 x}{x ^{3}} = - \frac{1}{6} .

The square root and logarithm are both real in a sufficiently small neighborhood of zero, so the two-sided real limit is well defined.

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