$0\times\infty$ and $1^\infty$

First consider

$\begin{aligned} A & = \lim_{n \to \infty} \left(1+\frac 1n \right)^n \\ & = \lim_{n \to \infty} \left(1 + \frac nn + \frac {n(n-1)}{2!n^2} + \frac {n(n-1)(n-2)}{3!n^3} + \cdots + \frac 1{n^n} \right) \\ & = \lim_{n \to \infty} \left(1 + 1 + \frac 1{2!}\left(1- \frac 1n\right) + \frac 1{3!}\left(1 - \frac 3n + \frac 2{n^2} \right) + \cdots + \frac 1{n^n} \right) \\ & = \frac 1{0!} + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \cdots \\ & = e \end{aligned}$

Since $A=e$ , the Euler's number , we have

$\begin{aligned} B & = \lim_{n \to \infty} n \left(\left(1+\frac 1n \right)^n - e \right) \\ & = \lim_{n \to \infty} n \left(1 + 1 + \frac 1{2!}\left(1- \frac 1n\right) + \frac 1{3!}\left(1 - \frac 3n + \frac 2{n^2} \right) + \cdots + \frac 1{n^n} - e \right) \\ & = \lim_{n \to \infty} n \left(-\frac 1{2!n} - \frac 1{3!}\left(\frac 3n - \frac 2{n^2} \right) - \frac 1{4!}\left(\frac 6n - \frac {11}{n^2} + \frac 6{n^3} \right) - \cdots \right) \\ & = \lim_{n \to \infty} \left(-\frac 1{2!} - \frac 1{3!}\left(3 - \frac 2n \right) - \frac 1{4!}\left(6 - \frac {11}n + \frac 6{n^2} \right) - \cdots \right) \\ & = - \frac 12 - \frac 12 - \frac 14 - \frac 1{12} - \cdots \\ & = -\frac 12 \left(\frac 1{0!} + \frac 1{1!} + \frac 1{2!} + \frac 1{3!} + \cdots\right) \\ & = - \frac e2 \end{aligned}$

Therefore, $\dfrac BA = \dfrac {-\frac e2}e = - \dfrac 12 = \boxed{-0.5}$