sin7x除以sin3x的极限
答:sin7x ~ 7x sin3x ~ 3x lim 7x/3x = 7/3
答:=lim sin(3x)/(3x) * lim (7x)/sin(7x) * (3x)/(7x)= (1)(1)(3/7)= 3/7
答:sin 3x/(3x)*7x/(sin7x)*3/7 (当趋于0)=3/7
答:3/7 当lim x-->0时 sin3x~3x sin7x~7x 所以结果就是3/7
答:解:lim(x->0)[sin(3x)/sin(7x)]=lim(x->0)[(3/7)*(sin(3x)/(3x))*((7x)/sin(7x))]=(3/7)*{lim(x->0)[sin(3x)/(3x)]}*{lim(x->0)[(7x)/sin(7x)]} =(3/7)*1*1 (应用重要极限lim(z->0)(sinz/z)=1)=3/7。
答:3/7 你学了无穷小的比较了么,有个等价无穷小概念 当x→0时,sinx~x,tanx~x,也就是说sinx和x是等价的,tanx和x也是等价的(仅x→0时有效)所以就可以化简为lim 3x/7x,因为x≠0约掉x得3/7
答:2、sin(3x)/2x=3/2*sin(3x)/3x lim(3/2)*sin(3x)/3x=3/2 lim sin(3x)/3x limsin3x/3x=1 所以上式极限=3/2 === 一般的极限可以直接代入求解;分式的话,如果是高项式,通常是最高次数项的系数比。如lim 7x^-6x+1/5x^+2x+1(x趋于无穷)=7/5 如果含无理式,则一般先进行分...
答:lim x趋近于0 sin3x~3x 结果=3/7
答:lim x趋近于0 sin3x~3x 结果=3/7
答:如图
网友评论:
邹田19459594558:
lim(sin3x/sin7x)的极限 -
29004訾昨
: x->0 lim sin(3x)/sin(7x) =lim sin(3x)/(3x) * lim (7x)/sin(7x) * (3x)/(7x) = (1)(1)(3/7) = 3/7
邹田19459594558:
(sin7x一sin3x)/x在x趋向0时的极限值 -
29004訾昨
:[答案] lim(x->0) (sin7x- sin3x)/x =lim(x->0) (7x- 3x)/x =4
邹田19459594558:
x趋向于零时 为什么sin7x除已sin3x等于7分之三 -
29004訾昨
: 当x趋向于0时,sin(7x)和sin(3x)可以用等价无穷小代换为7x和3x; 这样,x约掉后,结果为7/3
邹田19459594558:
求x趋于0时sin7x/sin3x的极限 -
29004訾昨
: 使用等价无穷小. x->0时 sin7x ~ 7x sin3x ~ 3x lim 7x/3x = 7/3
邹田19459594558:
lim(sin3x/sin7x)的极限 -
29004訾昨
:[答案] x->0 lim sin(3x)/sin(7x) =lim sin(3x)/(3x) * lim (7x)/sin(7x) * (3x)/(7x) = (1)(1)(3/7) = 3/7
邹田19459594558:
sinx*sin3x的积分怎么求 -
29004訾昨
: ∫[sinxsin(3x)]dx =∫½[cos(x-3x)-cos(x+3x)]dx =½∫[cos(-2x)-cos(4x)]dx =½∫[cos(2x)-cos(4x)]dx =½∫cos(2x)dx -½∫cos(4x)dx =¼∫cos(2x)d(2x)-⅛∫cos(4x)d(4x) =-¼sin(2x)+⅛sin(4x)+C提示:先对sinxsin(3x)进行恒等变形:积化和差.
邹田19459594558:
x - >1时(sin5x - sin3x)/sinx的极限是多少 -
29004訾昨
: 因为(sin5x-sin3x)/sinx在x=1的领域内连续,所以极限值等于函数值,答案为 (sin5-sin3)/sin1
邹田19459594558:
lim X趋近0 sin7X除以sin2X -
29004訾昨
: sin7x/sin2x=sin7x/2sinxcosx x→0,limsin7x/2sinxcosx=lim7x/2x=7/2x→∞lim(x^2-1)/(3x^2-x-1) =lim (1-1/x^2)/(3-1/x-1/x^2) =1/3
邹田19459594558:
极限sin5x除以sin7x -
29004訾昨
: lim<x→0>(sin5x/sin7x) =lim<x→∞>[(7/5)·((sin5x)/5x)/((sin7x)/7x)] =7/5.
邹田19459594558:
求当x趋近于π时sin4x÷sin7x的极限 -
29004訾昨
: 原式=limx→π (4*cos4x)/(7*cos7x),(0/0型,洛必塔法则求导)=4cos4π/7cos7π=-4/7.