jwbf.net
当前位置:首页 >> sin& x%sin3x >>

sin& x%sin3x

sin3x=sin(x+2x) =sinxcos2x+cosxsin2x =sinx[1-2(sinx)^2]+cosx(2sinxcosx) =sinx[1-2(sinx)^2]+2sinx(cosx)^2 =sinx-2(sinx)^3+2sinx[1-(sinx)^2] =sinx-2(sinx)^3+2sinx-2(sinx)^3 =3sinx-4(sinx)^3 所以 -sin3x=4(sinx)^3-3sinx.

3sin²x cosx-3cos3x 解析: y' =(sin³x-sin3x)' =3sin²x(sinx )'-cos3x*(3x)' =3sin²xcosx-3cos3x

令a=1/x 则原式=lim(a→0)[sin3a/a+asin(3/a)] 显然lim(a→0)sin3a/a =lim(a→0)=3a/a =3 而lim(a→0)asin93/a)中 3/a趋于无穷 所以sin(3/a)在[-,]震荡,即有界 无穷小乘有界还是无穷小 所以lim(a→0)asin93/a)=0 两部分极限都存在 所以原式=3+0=3

无穷小代换: 原式=lim (x-2x)/(x+3x) =lim -x/(4x) =-1/4

sin3x =sin(2x+x) =sin2xcosx+cos2xsinx =2sinxcosxcosx+cos2xsinx =2sinxcos²x+cos2xsinx sin3x-sin2x+sinx =2sinxcos²x+cos2xsinx-sin2x+sinx=0 若sinx=0 此时x=kπ 若sinx≠0时,得: 2cos^2x+cosx+1=0 相当于一元二次方程2x^2+x+1=0...

解法一:等价无穷小 lim sinx/sin3x x→0 =lim x/(3x) x→0 =⅓ 解法二:洛必达法则 lim sinx/sin3x x→0 =lim cosx/(3cos3x) x→0 =cos0/(3·cos0) =1/(3·1) =⅓

x->0 lim sin(3x)/sin(7x) =lim sin(3x)/(3x) * lim (7x)/sin(7x) * (3x)/(7x) = (1)(1)(3/7) = 3/7

等价无穷小,2/3

lim(x->0) (3sinx - sin3x) /x^3 (0/0) =lim(x->0) (3cosx - 3cos3x) /(3x^2) (0/0) =lim(x->0) (-3sinx + 9sin3x) /(6x) (0/0) =lim(x->0) (-3cosx + 27cos3x) /6 =24/6 =4

=-1/3*cos3x 换元积分,很简单的题

网站首页 | 网站地图
All rights reserved Powered by www.jwbf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com