由罗必塔法则得[ln(1+x³)]'/[ln(1+x²)]'=[3x²/(1+x³)]/[2x/(1+x²)]=(3x³+3x)/(2x³+2)=(3+ 3/x²)/(2+ 2/x³)x->+∞ 3/x²->0 2/x³->0(3+ 3/x²)/(2+ 2/x&#...