计算lim[/]^/2

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/19 02:05:11

计算lim[/]^/2

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[(1-y)/(1+y)]/y
=e^(1/2)lim[y→0] {2(1+8y)/(y²-1)+8ln[(1-y)/(1+y)]},←洛必达法则
=e^(1/2)*2(1)/(-1)+8ln(1)
=e^(1/2)*(-2)
=e^-1
=1/e

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