求极限:lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)如题,过程请详细!~~急在线等convoi 请问lim(n→∞)(1+1/n)^n=E 中n可以是小于0的么,它可以从哪里开始取?kjws12345 你的答案是错的,因为累乘只适用于
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/12 07:44:06
![求极限:lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)如题,过程请详细!~~急在线等convoi 请问lim(n→∞)(1+1/n)^n=E 中n可以是小于0的么,它可以从哪里开始取?kjws12345 你的答案是错的,因为累乘只适用于](/uploads/image/z/3937810-58-0.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90%EF%BC%9Alim%28n%E2%86%92%E2%88%9E%29%5B%EF%BC%883n%2B1+%EF%BC%89%2F%EF%BC%883n%2B2%EF%BC%89%5D%5E%28n%2B1%29%E5%A6%82%E9%A2%98%2C%E8%BF%87%E7%A8%8B%E8%AF%B7%E8%AF%A6%E7%BB%86%21%7E%7E%E6%80%A5%E5%9C%A8%E7%BA%BF%E7%AD%89convoi+%E8%AF%B7%E9%97%AElim%28n%E2%86%92%E2%88%9E%29%EF%BC%881%2B1%2Fn%EF%BC%89%5En%3DE+%E4%B8%ADn%E5%8F%AF%E4%BB%A5%E6%98%AF%E5%B0%8F%E4%BA%8E0%E7%9A%84%E4%B9%88%EF%BC%8C%E5%AE%83%E5%8F%AF%E4%BB%A5%E4%BB%8E%E5%93%AA%E9%87%8C%E5%BC%80%E5%A7%8B%E5%8F%96%EF%BC%9Fkjws12345+%E4%BD%A0%E7%9A%84%E7%AD%94%E6%A1%88%E6%98%AF%E9%94%99%E7%9A%84%EF%BC%8C%E5%9B%A0%E4%B8%BA%E7%B4%AF%E4%B9%98%E5%8F%AA%E9%80%82%E7%94%A8%E4%BA%8E)
求极限:lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)如题,过程请详细!~~急在线等convoi 请问lim(n→∞)(1+1/n)^n=E 中n可以是小于0的么,它可以从哪里开始取?kjws12345 你的答案是错的,因为累乘只适用于
求极限:lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)
如题,过程请详细!~~
急
在线等
convoi 请问lim(n→∞)(1+1/n)^n=E 中n可以是小于0的么,它可以从哪里开始取?kjws12345 你的答案是错的,因为累乘只适用于有限次运算
求极限:lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)如题,过程请详细!~~急在线等convoi 请问lim(n→∞)(1+1/n)^n=E 中n可以是小于0的么,它可以从哪里开始取?kjws12345 你的答案是错的,因为累乘只适用于
上式=lim(1-1/(3n+2))^-(3n+2)/-3 因为3n+2和3n+3是等价无穷大
由e的定义 上式=e^-(1/3)
原式=lim[1-1/(3n+2)]^(n+1)
由于n趋近正无穷,所以[1-1/(3n+2)]小于1
(-1,1)间的数的正无穷次方无限趋近于0 这个应该知道的
所以结果为0
极限值为零
lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)
=lim(n->无穷)(1+1/(3n+2))^[1/3(3n+2)-1]
=lim(n->无穷)[(1+1/(3n+2))^(3n+2)]^1/3/[(3n+1)/(3n+2)]
((3n+1),(3n+2)等价无穷大)
(重要极限:lim(n->无穷)(1+1/n)^n=e)
=lim(n->...
全部展开
lim(n→∞)[(3n+1 )/(3n+2)]^(n+1)
=lim(n->无穷)(1+1/(3n+2))^[1/3(3n+2)-1]
=lim(n->无穷)[(1+1/(3n+2))^(3n+2)]^1/3/[(3n+1)/(3n+2)]
((3n+1),(3n+2)等价无穷大)
(重要极限:lim(n->无穷)(1+1/n)^n=e)
=lim(n->无穷)[(1+1/(3n+2))^(3n+2)]^1/3/
(1-1/(3n-2))
=e^1/3/1
=e^1/3
收起