比较难的求极限题目(只要思路) 为什么lim(n趋向于无穷)1/n*{(n+1)(n+2).(n+n)}^(1/n)=4
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比较难的求极限题目(只要思路) 为什么lim(n趋向于无穷)1/n*{(n+1)(n+2).(n+n)}^(1/n)=4/e?
![比较难的求极限题目(只要思路) 为什么lim(n趋向于无穷)1/n*{(n+1)(n+2).(n+n)}^(1/n)=4](/uploads/image/z/18666836-44-6.jpg?t=%E6%AF%94%E8%BE%83%E9%9A%BE%E7%9A%84%E6%B1%82%E6%9E%81%E9%99%90%E9%A2%98%E7%9B%AE%EF%BC%88%E5%8F%AA%E8%A6%81%E6%80%9D%E8%B7%AF%EF%BC%89+%E4%B8%BA%E4%BB%80%E4%B9%88lim%EF%BC%88n%E8%B6%8B%E5%90%91%E4%BA%8E%E6%97%A0%E7%A9%B7%EF%BC%891%2Fn%2A%7B%28n%2B1%29%28n%2B2%29.%28n%2Bn%29%7D%5E%281%2Fn%29%3D4)
为了就算方便,令A=(1/n)[(n+1)(n+2)(n+3).(n+n)]^(1/n)
则 A=[(n+1)(n+2)(n+3).(n+n)/n^n]^(1/n)
={[(n+1)/n][(n+2)/n][(n+3)/n].[(n+n)/n]}^(1/n)
=[(1+1/n)(1+2/n)(1+3/n).(1+n/n)]^(1/n)
∴lnA=(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+.+ln(1+n/n)] (两边取自然对数 )
==>ln[lim(n->∞)A]=lim(n->∞)(lnA) (应用对数函数的连续性)
=lim(n->∞){(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+.+ln(1+n/n)]}
=∫(0,1)ln(1+x)dx (根据定积分定义得,符号∫(0,1)表示从0到1积分)
=[xln(1+x)]│(0,1)-∫(0,1)xdx/(1+x) (应用分部积分法)
=ln2-∫(0,1)[1-1/(1+x)]dx
=ln2-[x-ln(1+x)]│(0,1)
=ln2-(1-ln2)
=2ln2-1
=ln4-lne
=ln(4/e)
==>lim(n->∞)A=4/e (两边取反自然对数)
故 lim(n->∞){(1/n)[(n+1)(n+2)(n+3).(n+n)]^(1/n)}=4/e.
则 A=[(n+1)(n+2)(n+3).(n+n)/n^n]^(1/n)
={[(n+1)/n][(n+2)/n][(n+3)/n].[(n+n)/n]}^(1/n)
=[(1+1/n)(1+2/n)(1+3/n).(1+n/n)]^(1/n)
∴lnA=(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+.+ln(1+n/n)] (两边取自然对数 )
==>ln[lim(n->∞)A]=lim(n->∞)(lnA) (应用对数函数的连续性)
=lim(n->∞){(1/n)[ln(1+1/n)+ln(1+2/n)+ln(1+3/n)+.+ln(1+n/n)]}
=∫(0,1)ln(1+x)dx (根据定积分定义得,符号∫(0,1)表示从0到1积分)
=[xln(1+x)]│(0,1)-∫(0,1)xdx/(1+x) (应用分部积分法)
=ln2-∫(0,1)[1-1/(1+x)]dx
=ln2-[x-ln(1+x)]│(0,1)
=ln2-(1-ln2)
=2ln2-1
=ln4-lne
=ln(4/e)
==>lim(n->∞)A=4/e (两边取反自然对数)
故 lim(n->∞){(1/n)[(n+1)(n+2)(n+3).(n+n)]^(1/n)}=4/e.
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