Xn=[(n-1)/(n+1)]^n 求数列极限

网友回答

lim [(n-1)/(n+1)]^n
=lim [(n+1-2)/(n+1)]^n
=lim [1+(-2)/(n+1)]^n
=lim [1+(-2)/(n+1)]^(n+1-1)
=lim [1+(-2)/(n+1)]^(n+1) * [1+(-2)/(n+1)]^(-1)
=lim [1+(-2)/(n+1)]^(n+1) * lim [1+(-2)/(n+1)]^(-1)
=lim [1+(-2)/(n+1)]^(n+1) * 1
=lim [1+(-2)/(n+1)]^[(n+1)/(-2) * (-2)]
=lim {[1+(-2)/(n+1)]^[(n+1)/(-2)]}^(-2)
={lim [1+(-2)/(n+1)]^[(n+1)/(-2)]}^(-2)
根据重要的极限：lim (1+1/n)^n=e
=e^(-2)
有不懂欢迎追问
======以下答案可供参考======
供参考答案1:
＝lim﹛[1－2/﹙n＋1﹚]^﹙n＋1﹚/2﹜²/[1－2/﹙n＋1﹚]＝1/e²
供参考答案2:
Xn=[(n-1)/(n 1)]^n
=[1-2/(n 1)]^(-(n 1)/2)*(-2/(n 1))*n
底数趋于e,指数(-2n/(n 1))趋于-2
极限为e^(-2)