lim[x->+∞](x/(x-1))^√x=lim[x->+∞]e^(√xln(x/(x-1)))考察lim[x->+∞]√xln(x/(x-1))=lim[x->+∞]ln(x/(x-1))/(1/√x) (0/0型,可以用罗比塔法则)=lim[x->+∞](ln(x/(x-1)))'/(1/√x)'=lim[x->+∞]2√x/(x-1)=0所...