设 f(n) = [(a^1/n+b^1/n)/2]^n ,ln f(n) = n * ln[(a^1/n+b^1/n)/2]
令 t=1/n,n->+∞,t->0,ln f(n) = ln[(a^t + b^t)/2] / t
当t->0时,a^t -1 t * lna,b^t - 1 t * lnb
(a^t + b^t)/2 ->1,ln[(a^t + b^t)/2] (a^t + b^t)/2 -1
lim(n->∞) ln f(n)
= lim(t->0) [(a^t + b^t)/2 -1] / t
= (1/2) lim(t->0) [(a^t -1)/ t + (b^t -1)/ t]
= (1/2) (lna + lnb) = ln (ab)^(1/2)
原式 = (ab)^(1/2)