x→0时2x+x^3→0 x-x^2→0 即tan(2x+x^3)→0 ,sin(x-x^2)→0
分子分母同时→0 适用于洛必塔法则
lim(x→0)[tan(2x+x^3)/sin(x-x^2)]
=lim(x→0){[tan(2x+x^3)]'/[sin(x-x^2)] '}
=lim(x→0){[sec(2x+x^3)]^2*(2+3x^2)]/[cos(x-x^2) *(1-2x)]}
x→0,sec(2x+x^3)→1,cos(x-x^2)→1,(2+3x^2)→2 ,(1-2x)→1
所以上式极限为
=2