http://bbs.freekaoyan.com/thread-513074-1-1.html
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[(a^(1/n)+b^(1/n))/2]^n
=e^[nln(a^(1/n)+b^(1/n))/2]
~e^[n(a^(1/n)+b^(1/n))/2-1]
=e^1/2[(a^(1/n)-1)/(1/n)+(b^(1/n)-1)/(1/n)](n-> ÎÞÇî)
=e^1/2[(a^t-1)/t+b^t-1)/t](Áît=1/n,t->0),
¿¼ÂǺ¯Êýf(x)=a^xºÍg(x)=b^x,
ÓÉÀ¸ñÀÊÈÕ¶¨Àí£¬´æÔÚ0<u<t,0<v<tʹµÃ(a^t-1)/t=(a^t-a^0)/t=f'(u)=a^u*lna->lna(t->0ʱ£¬u->0),
ͬÀí£¬(b^t-1)/t=(b^t-b^0)/t=g'(v)=b^v*lnb->lnb(t->0ʱ£¬v->0),
ËùÒÔÓÐe^1/2[(a^t-1)/t+b^t-1)/t]->e^1/2(lna+lnb)=e^1/2*lnab=sqrt(ab)
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