作者keith291 (keith)
看板trans_math
标题Re: [考古]help me
时间Tue Oct 12 23:03:39 2010
※ 引述《orlando1988 (orlando)》之铭言：
: lim(1-1/2n)^3n =??
: n->OO
令y=(1-1/2n)^3n lny= ln(1 - 1/2n)/(1/3n)
1/{(2n^2)(1 - 1/2n)}
lim lny = lim ----------------------- = -3/2 = ln( lim y ) (因y=lnx在x>0连续)
n→∞ n→∞ -1/3n^2 n→∞
得
lim(1-1/2n)^3n = e^(-3/2)
n→∞
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◆ From: 61.231.103.234
1F：→ orlando1988:看无ˇˇ 203.64.181.205 10/13 16:49
2F：推 a88241050:没必要写那麽复杂吧..我也看不太懂111.248.224.129 10/13 18:08
3F：→ a88241050:lim(1+1/n)^n(n->oo)=e111.248.224.129 10/13 18:10
4F：→ a88241050:＞用这个代就好了111.248.224.129 10/13 18:10
5F：→ keith291:用到L'hospital rule一次218.166.108.120 10/13 18:30
6F：→ keith291:然後不用楼楼上那个作法的原因是因为218.166.108.120 10/13 18:30
7F：→ keith291:变数变换後还要说明趋近"负无限"和原本218.166.108.120 10/13 18:31
8F：→ keith291:定义中的正无限会得一样结果 有点麻烦218.166.108.120 10/13 18:31
※ 编辑: keith291 来自: 218.166.108.120 (10/13 18:33)
9F：推 arthur104:他...应该写的很简洁了吧XD 123.193.37.84 10/13 18:38