不难,可以取对数lim[x→0] x^x=e^lim[x→0] ln(x^x)=e^lim[x→0] xlnx ,这里取y=1/x则x=1/y,当x→0,y→∞=e^lim[y→∞] (1/y)ln(1/y)=e^lim[y→∞] -(lny)/y ,运用洛必达法则求导得=e^-lim[y→∞] 1/y ,取极限=e^0=1...