X->0 (sinx)^2/(1-cosx+sinx) 的极限,

方法一：0/0型极限,用L'Hospital法则
lim(x→0) sin²x/(1-cosx+sinx)
=lim(x→0) (sin²x)'/(1-cosx+sinx)'
=lim(x→0) (2sinxcosx)/(sinx-cosx)
=0
方法二：
lim(x→0) sin²x/(1-cosx+sinx)
=lim(x→0) [2sin(x/2)cos(x/2)]²/[2sin²(x/2)+2sin(x/2)cos(x/2)]
=lim(x→0) [2sin(x/2)cos²(x/2)]/[sin(x/2)+cos(x/2)]
=0我看参考书上是上下同除以X^2 然后等价无穷小，等于4/3加减是不能用等价无穷小的吧谢谢“恋任世纪”的评论(1-cosx+sinx)'=sinx+cosx笔误