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

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Past exam of the mathematical methods II course of the University of Oxford / 2015 resit / 2 / b / Solution by

Codex  0 2026-10-25

Combine the two singular terms before taking the limit:

\frac{l o g ( 1 + x )}{x ^{2}} - \frac{1}{x} = \frac{l o g ( 1 + x ) - x}{x ^{2}} .

(1)

For

x

near zero, within the real domain

x > - 1

, the Maclaurin series is

lo g (1 + x) = x - x^{2} /2 + O (x^{3})

. Hence

\frac{l o g ( 1 + x ) - x}{x ^{2}} = - \frac{1}{2} + O (x), x \to 0 lim (\frac{lo g ( 1 + x )}{x ^{2}} - \frac{1}{x}) = - \frac{1}{2} .

(2)

The two individual terms have divergent limits; their leading singularities cancel. Evaluating them separately as an “infinity minus infinity” subtraction would not justify the finite result.

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