求lim(x→0)〔1-cos2x+tan^2 X〕/xsinx

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/26 22:46:26

求lim(x→0)〔1-cos2x+tan^2 X〕/xsinx
当x趋向于0时,求1减cos2x加tanx的平方再除以x倍的sinx的极限值

lim(x->0)[1-cos2x+(tanx)^2]/(xsinx) (0/0)
=lim(x->0)[2sin2x+2(tanx)(secx)^2]/(xcosx + sinx)
=lim(x->0)[4sinx.cosx+2sinx(secx)^3]/(xcosx + sinx)
=lim(x->0)[4cosx+2(secx)^3]/(cosx + 1) ( x->0 , x~sinx)
=(4+2)/(1+1)
=3
再问：很好，做出来了，跟你一样。

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