 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/19/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

A > B

e^{A / x^{2}}

. The expression becomes

(A + B e^{- (A - B) / x^{2}}) / (1 + e^{- (A - B) / x^{2}})

, which tends to

A

since

A - B > 0

. If

B > A

, the same argument gives

B

. If

A = B

, the expression equals

A

for every

x \neq = 0

. Thus

x \to 0 lim \frac{A e ^{A / x^{2}} + B e ^{B / x^{2}}}{e ^{A / x^{2}} + e ^{B / x^{2}}} = max (A, B) .

This is the two-term case of exponential concentration of a weighted mean, and holds for negative as well as positive parameters.

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