第1个回答 2012-11-30
lim(x→+∞)[(1/x)*(a^x-1)/(a-1)]^(1/x)
=lim(x→+∞)(1/x)^(1/x)*lim(x→+∞)[(a^x-1)/(a-1)]^(1/x)
lim(x→+∞)ln[(1/x)^(1/x)]
=lim(x→+∞)(1/x)ln(1/x)
=lim(x→+∞)-lnx/x
=0
lim(x→+∞)(1/x)^(1/x)=1
lim(x→+∞)ln{[(a^x-1)/(a-1)]^(1/x)}
=lim(x→+∞)(1/x)ln[(a^x-1)/(a-1)]
=lim(x→+∞)ln[(a^x-1)/(a-1)]/x
=lim(x→+∞)ln(a^x-1)/x - lim(x→+∞)ln(a-1)/x
=lna
lim(x→+∞)[(a^x-1)/(a-1)]^(1/x)=a
原式=1*a=a本回答被提问者采纳