 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/19/b/solution

Use the Maclaurin series on the positive side:

f (x) = \frac{x ^{2}}{2} + O (x^{4}), g (x) = \frac{x ^{3}}{6} + O (x^{5}), h (x) = x [1 + O (x)] .

In the last expression the original square root covers

e^{x} - 1

, whose leading term is

x > 0

. Hence

g / f \to 0

and

f / h \to 0

. All three are positive for sufficiently small positive

x

, so

g (x) < f (x) < h (x) for sufficiently small x > 0.

Thus

h

is greatest and

g

least in the stated limit.

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