 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/2/a/solution

sin x = x - x^{3} /6 + O (x^{5})

cancels the numerator's linear term, leaving

\frac{s i n x - x}{x ^{3}} = - \frac{1}{6} + O (x^{2}) .

Thus

x \to 0 lim (sin x - x) / x^{3} = - 1/6

. The same expansion applies for positive and negative

x

, so this is a two-sided limit.

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