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求极限 (tanx-x)/(x-sinx) x→0时的极限
人气:494 ℃ 时间:2019-08-21 09:20:50
解答
lim(x→0)[(tanx-x)/(x-sinx)]
=lim(x→0)[(-x+tanx)/(x-sinx)],0/0型,应用洛必达法则
=lim(x→0)[(-1+(1/cos²x))/(1-cosx)]
=lim(x→0)[(1-cos²x)/(cos²x(1-cosx))]
=lim(x→0)[sin²x/(cos²x-cos³x)],0/0,再用洛必达法则
=lim(x→0)[2sinxcosx/(3sinxcosx-2sinxcosx)]
=lim(x→0)(2sinxcosx/sinxcosx)
=lim(x→0)2
=2
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