Question

高数：limx-π/2(2sinx^2-sinx-1)/(sinx^2+sinx-2)求极限

Answer 1

可以设t=sinx,因为x-π/2 所以 t-1

用复合函数的积分法则代换得：limt-1(2t^2-t-1)/(t^2+t-2)

其中：(2t^2-t-1)/(t^2+t-2) = (2t+1)(t-1)/(t+2)(t-1) = (2t+1)/(t+2)

limt-1(2t^2-t-1)/(t^2+t-2) = limt-1(2t+1)/(t+2)=3/3=1