求lim n趋向于无穷 n(n^1/n -1)/ln n
1个回答

令t=1/n 则:t→0

lim n(n^1/n -1)/ln n

=lim﹣(1/t^t -1)/tlnt

=lim(1- 1/t^t )/lnt^t

∵t→0

∴可求 t→0时,t^t 极限为 1

令x=t^t,则x→1

∴原始化为:

lim(1- 1/x)/lnx (0/0型,用罗比达法则)

=lim(1- 1/x)'/(lnx)'

=lim(1/x²)/(1/x)

=lim1/x

=1

综上,原式极限为 1