> 数学 >
求极限lim(x→0)(tanx-sinx)/sin³x
人气:204 ℃ 时间:2020-03-21 11:57:23
解答
tanx-sinx=sinx/cosx-sinx
=[sinx(1-cosx)]/cosx
(tanx-sinx)/sin3x=(1-cosx/cosx)sin2x
=(1-cosx/cosx)/1-cos2x
=(1-cosx/cosx)/[(1-cosx)(1+cosx)]
=1/[cosx(1+cosx)]
lim趋向于0时应该=1/2
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版