求极限limx->0时,(tanx-sinx)/(1-cos2x)的极限,
一楼二楼都不对,和答案不一样呢
人气:406 ℃ 时间:2020-04-06 04:02:16
解答
(tanx-sinx)/(1-cos2x)= (tanx-sinx)/(1-1+2(Sinx)^2)= (tanx-sinx)/(sinx)^2= ((tanx-sinx)cotx)/(cotx(sinx)^2)= (1-cosx)/(sinxcosx)用洛必大法则,(1-cosx)'/(sinxcosx)'= sinx/(2(cosx)^2-1)= 0/(2-1)= 0
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