limn趋向于无穷(n+1)(2的n次方+1)⼀2的n+1次方+1

2025-03-09 21:01:10
推荐回答(1个)
回答1:

lim [(n+1)(2^n+1)]/[2^(n+1)+1] [ 分子分母同乘以2^(-n) ]
= lim {(n+1)[1+2^(-n)]}/[2+2^(-n)] = ∞