Question

[(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

Answer 1

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)