【(sinx)^2-x^2cosx^2】/(sinx)^4中x趋向于0时的极限

lim(x→0) [(sinx)^2-x^2(cosx)^2]/(sinx)^4
(sinx)^2 =(1-cos2x)/2 x^2(cosx)^2=x^2(1+cos2x)/2
=2 lim(x→0) [(1-cos2x)-x^2(1+cos2x)]/(1-cos2x)^2
=2lim(x→0)(1-x^2-x^2cos2x-cos2x)'/(1-2cos2x+(cos2x)^2)'
=2lim(x→0)(-2x-2xcos2x+2x^2sin2x+2sin2x)/(4sin2x-4sin2xcos2x)
=lim(x→0)(-x-xcos2x+x^2sin2x+sin2x)' /(sin2x-sin2xcos2x)'
=lim(x→0)(-1-cos2x+4xsin2x+2x^2cos2x+2cos2x)'/(2cos2x-2cos4x)'
=lim(x→0)(6sin2x+12xcos2x-4x^2sin2x-4sin2x)'/(4sin2x+8sin4x)'
=lim(x→0)(24cos2x-32xsin2x-8x^2cos2x-8cos2x)' / (8co2x+32cos4x)'
=(24-8)/(8+32)=16/40=2/5先把cosx^2换为1-sinx^2，分出常量1，再变为{（sinx-x)/x^3}*{(sinx+x)/x}+1的极限。前者洛必达，后者无穷小代换，即为（-1/6)*2+1=2/3，我同学告诉我的，麻烦了。