 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017/3/b/solution

For real

x

sufficiently close to zero,

1 + x > 0

. Taking the logarithm and using its Maclaurin series gives

lo g [(1 + x)^{1/ x}] = \frac{l o g ( 1 + x )}{x} = 1 - \frac{x}{2} + \frac{x ^{2}}{3} + O (x^{3}) ⟶ 1.

Continuity of the exponential then gives

x \to 0 lim (1 + x)^{1/ x} = e .

This is a two-sided real limit on the punctured neighbourhood where the base is positive.

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