> 数学 >
求极限lim(x趋向于0) (x-tanx)/xsinx^2
人气:417 ℃ 时间:2020-03-28 09:13:41
解答
lim(x-tanx)/xsinx^2
=lim(x-tanx)/(x * x^2 * sinx^2 / x^2)等价无穷小量:
=lim(x-tanx)/(x * x^2)
=lim(1/x^2 - sinx/x * 1/[x^2 * (cosx)]
=lim(1 - secx)/x^2 罗必塔法则:
=lim(-tanxsecx)/2x 罗必塔法则:
=-1/2limsecx(tan^2 x + sec^2 x)
=-1/2我觉得你第三个等号时算错了,请修改...没算错啊,lim(x-tanx)/(x * x^2)=lim1/x^2 - sinx/x * 1/(x^2 * cosx)把tanx=sinx/cosx我只算到(xcosx-sinx)/x^2*cosx,算不到你的第三个=号那条式,请讲解一下(x-tanx)/(x * x^2) =x/(x * x^2) - tanx/(x * x^2) =1/x^2 - (sinx/cosx)/(x * x^2) =1/x^2 - sinx/x * 1/(cosx * x^2)
推荐
猜你喜欢
© 2025 79432.Com All Rights Reserved.
电脑版|手机版