作业帮大一求极限题求解[(a^1/n+b^1/n)/2]^n,n趋于无穷
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大一求极限题求解[(a^1/n+b^1/n)/2]^n,n趋于无穷
大一求极限题求解
[(a^1/n+b^1/n)/2]^n,n趋于无穷
大一求极限题求解[(a^1/n+b^1/n)/2]^n,n趋于无穷
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)
原式=lim(n->∞){e^[nln((a^(1/n)+b^(1/n))/2)]}
=lim(n->∞){e^[ln((a^(1/n)+b^(1/n))/2)/(1/n)]}
=lim(m->0){e^[ln((a^m+b^m)/2)/m]} (令m=1/n)
=e^{lim(m->0)...
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原式=lim(n->∞){e^[nln((a^(1/n)+b^(1/n))/2)]}
=lim(n->∞){e^[ln((a^(1/n)+b^(1/n))/2)/(1/n)]}
=lim(m->0){e^[ln((a^m+b^m)/2)/m]} (令m=1/n)
=e^{lim(m->0)[ln((a^m+b^m)/2)/m]} (应用初等函数的连续性)
=e^{lim(m->0)[ln((a^m*ln│a│+b^m*ln│b│)/2)]} (0/0型极限，应用罗比达法则)
=e^[(ln│a│+ln│b│)/2]
=e^ln(√(ab))
=√(ab)。
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