方法一：0/0型極限,用L'Hospital法則
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我看參考書(shū)上是上下同除以X^2 然后等價(jià)無(wú)窮小，等于4/3加減是不能用等價(jià)無(wú)窮小的吧謝謝“戀任世紀(jì)”的評(píng)論(1-cosx+sinx)'=sinx+cosx筆誤