a^(1/n)-1=bn
lna/n=ln(bn+1)
n(a^(1/n)-1)=lna*bn/ln(bn+1)
当n足够大时
0存在唯一kn使得1/kn<=b<=1/(kn+1)
那么由于1/(n+1)1/(kn+2)kn<1/ln(bn+1)1<-kn/(kn+1)1
那么lim bn/ln(bn+1)=1
故lim n(a^(1/n)-1)=lim lna*bn/ln(bn+1)=lna