Question

求lim[ntan(1/n)]^n^2的极限 ,n趋向无穷,最好用洛必达法则来求 答案是e^(1/3)

Answer 1

[ntan(1/n)]^n^2=e^{n^2ln[ntan(1/n)]}
又tan(1/n)和1/n是等价无穷小,所以lim ntan(1/n)=1
所以lim ln[ntan(1/n)]=0
所以构成不定型
由于f(n)是f(x)的子列,故把n换为x,若f(x)有极限,则f(n)也有极限
原式
lim n^2ln[ntan(1/n)]=lim x^2ln[xtan(1/x)]=lim [lnx+lntan(1/x)]/(1/x^2)
换元t=1/x,t→0
原式=lim [-lnt+lntant]/t^2
洛必达法则
=lim [-1/t+1/(sintcost)]/(2t)
=lim (t-sintcost)/(2t^2sintcost)
又lim cost=1,sint是t的等价无穷小,因此
=lim (t-sintcost)/(2t^3)
洛必达法则
=lim (1-cos2t)/(6t^2)=lim (sint)^2/3t^2=1/3
所以原极限为e^(1/3)