cool limit

Since $\frac{\left(1+\frac{1}{n}\right)^n}{\left(1-\frac{1}{n}\right)^n}=\left(1+\frac{2}{n-1}\right)^n,$ we let $u=\frac{1}{n-1}\Rightarrow n=\frac{1}{u}+1$ so that $\begin{aligned} \left(1+\frac{2}{n-1}\right)^n&=&(1+2u)^{\frac{1}{u}+1}\\ &=&(1+2u)\left(\exp\left(\frac{1}{u}\ln(1+2u)\right)\right)\\ &=&(1+2u)\left(\exp\left(2-2u+\frac{8}{3}u^2+\mathcal{O}(u^3)\right)\right) &&{\color{#3D99F6}(\text{Maclaurin expansion of }\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\mathcal{O}(x^4))}\\ &=&(1+2u)\left(e^2\times e^{-2u}\times e^{\frac{8}{3}u^2}\times\exp(\mathcal{O}(u^3))\right)\\ &=&(1+2u)\left(e^2(1-2u+2u^2)(1+\frac{8}{3}u^2)+\mathcal{O}(u^3)\right) &&{\color{#3D99F6}(\text{Maclaurin expansion of }e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\mathcal{O}(x^4))}\\ &=&e^2\left(1+\frac{2u^2}{3}+\mathcal{O}(u^3)\right). \end{aligned}$ Substituting back the definition of $u=\frac{1}{n-1}$ , the result follows.