作业帮计算lim[/]^/2
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/26 06:40:40
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
=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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