 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-resit/9/b/solution

For

∣ x ∣ < 1

(1 + x)^{- 1} = \sum_{j = 0}^{\infty} (- x)^{j}

. Integrating from zero to

x

ln (1 + x) = j = 1 \sum \infty \frac{( - 1 ) ^{j + 1} x ^{j}}{j} = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots .

On the real axis the series also converges at

x = 1

ln 2

, but diverges at

x = - 1

For fixed real

a

and sufficiently large positive

x

, the base in the requested limit is positive. Apply the series with argument

- a^{2} / x^{2}

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

Continuity of the exponential function therefore gives

x \to \infty lim (1 - \frac{a ^{2}}{x ^{2}})^{x^{2}} = e^{- a^{2}} .

This includes

a = 0

, when the expression is identically one.

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