利用定积分的定义求极限
方法:如果baf(x)dx存在,则
bbanbalimf(ak)f(x)dxannk1n
例15求极限
n22nn4kk1(1)
limn解:
n1nlim2lim2nnnn4kk1k1n1k14()2n1actan2x1actan2dx|0014x2221
nx2k2nnk1(2)
limn解:
1nx2k1nklimlim[x2()](x2t)dtx120nnnnnk1k1
n(3)
lim1nn(n1)(n2)(2n1)nn
解:因为
k0ln(1n)n1k由于
故
1nn(n1)(n2)(2n1)enn
lim1nnnln(1k)k1n10ln(1x)dx2ln21ln4e4lim1nnn(n1)(n2)(2n1)elne4ne