题目

计算lim[/]^/2

提问时间：2021-02-18

答案

lim[x→∞] [(x-1)/(x+1)]^[(x+8)/2]=e^(1/2)lim[x→∞] (x+8)ln[(x-1)/(x+1)]令x=1/y=e^(1/2)lim[y→0] (1/y+8)ln[(1/y-1)/(1/y+1)]=e^(1/2)lim[y→0] (1+8y)/y*ln{[(1-y)/y]/[(1+y)/y]}=e^(1/2)lim[y→0] (1+8y)ln...