x→+∞,arctanx→π/2,e^-x→0,(e^x+1)/(e^x-1)=(1+e^-x)/(1-e^-x)→1原式极限为π/2x→-∞,arctanx→-π/2,e^x→0,(e^x+1)/(e^x-1)=→-1原式极限为π/2综上,lim[x→∞] (e^x+1)/(e^x-1)arctanx→π/2希望可以帮到你...