[(a^1/n+b^1/n)/2]^n(a>0,b>0)在n趋于无穷时的极限,不用洛比达法则,只用重要极限和无穷小等价代

昨天做过
lim{[a^(1/n)+b^(1/n)]/2}^n
=e^lim n*ln{[a^(1/n)+b^(1/n)]/2}
而ln{[a^(1/n)+b^(1/n)]/2}=ln{1+{[a^(1/n)+b^(1/n)]/2}-1}~{a^(1/n)+b^(1/n)]/2}-1
原式化为 e^lim (n/2)*[a^(1/n)-1+b^(1/n)-1]
而a^x-1=xlna+o(x)
所以a^(1/n)-1+b^(1/n)-1=(1/n)*(lna+lnb)+o(1/n)
原式又化为 e^lim (n/2)*[a^(1/n)-1+b^(1/n)-1]
=e^lim(n/2)*[(1/n)*(lnab)+o(1/n)]
=e^lim ln(ab)^(1/2)
=(ab)^(1/2)