lim[1/(1*4)+1/(4*7).+1/(3n-1)(3n+1)]我怎么觉得题目有错阿,
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 18:03:29
xRMo@+{Ɩi\LHܐ X!I9ˈuCK"h4Mhmp>/
[Vٙ7ެU/?}bfqs%MόZM*I:t{z۩8ꤣ[lZcuo~(mt>:3-c&+,3[̡I4At)-EQbhUgHV쯨C/FwwɃ9mvy`3qvS{F,*,,_F3oV|u}qy
B҅zZF;4bOiT0*
=n[{,O>xХGHSK|U 7{;{5$$g|£p!
B.FHEz0?k͡YrUQpu=`P"n]>vBi(_غiu/Ȣ
lim(2an+4bn)=1 lim(3an-bn)=2 求lim(an+bn)
lim(2an+4bn)=1 lim(3an-bn)=2 求lim(an+bn)
lim[x-1%x+1]^[+4]
{An}{Bn},若lim(3An+Bn)=8,lim(6An-Bn)=1求lim(4An-Bn)与lim(AnBn)的值
lim[(4+7+...+3n+1)/(n^2-n)]=
计算下列极限:1)lim(n→∞) 1/n3 2)lim(n→∞)4n+1/3n-11)lim(n→∞) 1/n3 2)lim(n→∞)4n+1/3n-13) lim(n→∞) (1/3)n4)lim(n→∞)n3+2n-5/5n3-n 5) lim(n→∞)(1+1/2n)n 6) lim(n→∞)2x3-x2+1/3x2+2x-9 7) lim(x→0 )sin3x/sin7x8)
数学难题,求指教~~极限到底是什么玩意我还不懂.1.lim(x^2-2x+6)2.lim(x^2-4)/(x-2)3.lim(x^2-6x=8)/(x^-5x=4)4.lim[1/(1-x)-1/(1-x^3)5.lim(sin2x)/x^26.lim(sin5x)/x7.lim(tan3x)/(sin7x)8.lim[(1+x)/x]^2x例如这道例题~lim[(1/3)x+1]
lim(x)lim(x→∞) [(x-1)/(x+1)]^2x 为什么 lim(x->∞) {[(-2)/(x+1)]*2x} = -4
已知lim(3an+ 4bn)=8,lim(6an-bn)=1,则lim(2an +bn)的值是多少?
lim(7-1/n)=?是不是可以把7和1/n单独分开看,就是lim(an)=7,lim(bn)=lim(1/n)=0,lim(an-bn) =lim(7-1/n)=7,或者lim(bn)=lim(-1/n)=0,那么lim(an+bn)=lim(7+(-1/n))=lim(7-1/n)=7所以lim(7-1/n)
lim x-1/x平方+3x-4 x-1
lim(x趋于派/4)1+sin2x/1-cos4x=
lim[4/(4-x2) - 1/(2+x)]x∝-2
lim(n+3)(4-n)/(n-1)(3-2n)
lim(x->2) {(3x+16)^1/2-4}/x
lim x→∞(2x+1)/(3x-4)
lim(n^3+n)/(n^4-3n^2+1)
a=-4 b=-1 lim求和 如图