X->0 (sinx)^2/(1-cosx+sinx) 的极限,

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X->0(sinx)^2/(1-cosx+sinx)的极限,X->0(sinx)^2/(1-cosx+sinx)的极限,X->0(sinx)^2/(1-cosx+sinx)的极限,方法一：0/0型极限

X->0 (sinx)^2/(1-cosx+sinx) 的极限,
X->0 (sinx)^2/(1-cosx+sinx) 的极限,

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

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