limx+0x平方sinx分之一

别月疤2822limx的平方/sin方的x/3x趋向于0 -
湛肺柔18823913506 ______[答案] 当x-〉0时,sinx-〉x lim[x^2/sin^2(x/3)]=limx^2/(x/3)^2=1/9 x->0 x->0 或先用L'Hospital法则,即分式的分子和分母各自求导 lim(x^2)'/[sin^2(x/3)]'=lim[2x/[sin(x/3)*1/3]=lim2x/(x/9)=1/9 x->0

别月疤2822数学limx→0(x^3+1)/(x+1)sin2x -
湛肺柔18823913506 ______ =limx→0(x平方-x+1)/sin2x.limx→0sin2x/(x平方-x+1)=0.所以limx→0(x平方-x+1)/sin2x=无穷大.极限为无穷大.

别月疤2822limx→0x/Sin(x/2)的极限是什么 -
湛肺柔18823913506 ______[答案] limx→0x/Sin(x/2) =2limx→0(x/2)/Sin(x/2) =2*1 =2

别月疤2822limx趋近于0 x2/sin2(x/3) -
湛肺柔18823913506 ______ lim(x->0) x^2/(sin(x/3))^2 (0/0)=lim(x->0) 2x/sin(2x/3) (0/0)=lim(x->0) 2/[(2/3)cos(2x/3)]= 3

别月疤2822当x趋近于0时x的平方乘sin1/x的极限的求法高等数学第六版上册习题1 - 5,3(1)题 -
湛肺柔18823913506 ______[答案] sin1/x是有界量 所以sin1/x取值范围是[-1,1] 当x->0时 x^2->0,-x^2->0 所以 x^2*(-1)(-1是sin1/x的最小值)

别月疤2822求极限x趋近于0时,x平方除以正弦平方5分之x -
湛肺柔18823913506 ______ lim(x->0) x^2/[sin(x/5)]^2 =lim(x->0) x^2/(x/5)^2 =25

别月疤2822求极限limxˇ趋近0x平方分之cosx - 1 -
湛肺柔18823913506 ______[答案] 解法一:lim(x->0)[(cosx-1)/x^2] =lim(x->0)[-2(sin(x/2))^2/x^2] (应用倍角公式) =(-1/2){lim(x->0)[sin(x/2)/(x/2)]}^2 =(-1/2)*1^2 (应用重要极限lim(z->0)(sinz/z)=1) =-1/2; 解法二:lim(x->0)[(cosx-1)/x^2] =lim(x->0)[-sinx/(2x)] (0/0型极限...

别月疤2822什么情况下用两个重要极限求解,什么时候用无穷小量的性质求解 如题:limx→0xsin(1/x)=,用前者得1 后得0 -
湛肺柔18823913506 ______[答案] 你这题只能用一个公式来求,因为两个公式的成立条件不同,书上有的,自己看,这题等于0,用罗比达法则做.

别月疤2822limx趋于0.(sin x平方 +x )/(cos x的平方 - x)的极限,拜托了很急啊!如题,谁能告诉我哈,很急啊! -
湛肺柔18823913506 ______[答案] lim(x→0)(sin^2x + x)/(cos^2x - x) 因为 (sin^2x + x)/(cos^2x - x)在x=0出连续 所以lim(x→0)(sin^2x + x)/(cos^2x - x) =(sin0+0)/(cos0-x) =0+0/1-0 =0

别月疤2822求极限,lim x→0 sin(x - 2)÷x平方 - 4 lim x→正无穷(1+1÷3x) limx→...求极限,lim x→0 sin(x - 2)÷x平方 - 4 lim x→正无穷(1+1÷3x) limx→正无穷(1 - 2x)x分之... -
湛肺柔18823913506 ______[答案] lim(x→0) sin(x - 2)/(x² - 4)= lim(x→0) sin(x - 2)/[(x - 2)(x + 2)]= lim(x→0) sin(x - 2)/(x - 2) • lim(x→0) 1/(x + 2)= 1 • 1/(0 + 2)= 1/2lim(x→+∞) [1 + 1/(3x)]= 1 + lim(x→+∞) 1...