我用了洛必达法则计算:
lim(x→0) (sin²x-x²cos²x)/(x²sin²x)
=lim(x→0) (sinx-xcosx)(sinx+xcosx)/(x²sin²x)
=lim(x→0) (sinx+xcosx)x*(sinx-xcosx)/x³*(x/sinx)²
=lim(x→0) (sinx/x+cosx)*lim(x→0) (x/sinx)²*lim(x→0) (sinx-xcosx)/x³
=(1+1)*1*1/3
=2/3
其中:
lim(x→0) (sinx-xcosx)/x³
=lim(x→0) [cosx-(cosx-xsinx)]/(3x²),洛必达法则
=(1/3)lim(x→0) (xsinx/x²)
=(1/3)lim(x→0) sinx/x
=1/3
PS:代换的确不能在加减中应用,在乘除中可用,所以提出的共因素能用