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

If $A>B$, divide numerator and denominator by $e^{A/x^2}$. 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 $x\ne 0$. ThusThis is the two-term case of exponential concentration of a weighted mean, and holds for negative as well as positive parameters.