作业帮lim(1-1/n)^(n^2)=?
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/11/07 17:55:46
lim(1-1/n)^(n^2)=?
没有说明n趋向0或n趋向∞,那两种情况都写吧:
lim[n→0] (1-1/n)^(n²)
=e^lim[n→0] n²ln(1-1/n),令n=1/y,∴n²=1/y²
=e^lim[y→∞] ln(1-y)/y²,用洛必达法则
=e^lim[y→∞] [-1/(1-y)]/(2y)
=e^(-1/2)lim[y→∞] 1/[y(1-y)],分母趋向∞,于是整式趋向0
=e^(-1/2)*0
=e^0
=1
lim[n→∞] (1-1/n)^(n²)
=lim[n→∞] [1+(1/-n)]^[(-n)*lim[n→∞] (-n)]
=e^{-lim[n→∞] n}
=e^-∞
=1/e^∞
=1/∞
=0
根据e的定义 lim[n→∞] (1-1/n)^n=e,而上面的n=-n
lim[n→0] (1-1/n)^(n²)
=e^lim[n→0] n²ln(1-1/n),令n=1/y,∴n²=1/y²
=e^lim[y→∞] ln(1-y)/y²,用洛必达法则
=e^lim[y→∞] [-1/(1-y)]/(2y)
=e^(-1/2)lim[y→∞] 1/[y(1-y)],分母趋向∞,于是整式趋向0
=e^(-1/2)*0
=e^0
=1
lim[n→∞] (1-1/n)^(n²)
=lim[n→∞] [1+(1/-n)]^[(-n)*lim[n→∞] (-n)]
=e^{-lim[n→∞] n}
=e^-∞
=1/e^∞
=1/∞
=0
根据e的定义 lim[n→∞] (1-1/n)^n=e,而上面的n=-n
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