数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/14 15:02:18
x){6uӎ59A!'3W:Z8/H;9OPS_#Nӌ5ɫ5y1@]Iɶ6IEƀE3KΆj"2B!Bz]I&L+:^
fY MPf~T募nfB1XL547/.H̳
3gk\4ٜΗ`Og/xc_8Bhn"č
= Ox:g= wك
数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
数列极限
lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5
lim {[(3n^2+cn+1)-4n(an^2+bn)]/(an^2+bn)}=5
lim {[-4an^3+(3-4b)n^2+cn+1]/(an^2+bn)}=5
所以
-4a=0
3-4b=0
c/b=5
解得:
a=0
b=3/4
c=15/4
a=0
b=3/4
c=15/4
此题有问题
因为lim [(3n^2+cn+1)/(an^2+bn)]=3/a
而 lim (-4n)不存在
数列极限 lim {[(3n^2+cn+1)/(an^2+bn)]-4n}=5,则常数a=?b=?c=?
lim n →∞ (1^n+3^n+2^n)^1/n,求数列极限
数列极限计算lim(7n+4)/(5-3n)
数列的极限计算lim(3n²+4n-2)/(2n+1)²
数列极限例题lim(2n+1)/(3n-1)n→∞
高数 数列极限lim(1+ 2^n + 3^n)^(1/n) n趋于无穷大求极限
数列极限基本题已知数列{an}的极限为0,且有lim[(3n-2)an]=6,则lim[n(an)]=?
证明数列的极限证明lim(3n+1)/(2n+1)=3/2
根据数列极限的定义证明:lim(3n+1)/(2n+1)=3/2
求下列数列的极限:lim (2+3^n)/(1+3^(n+1))
求数列极限lim(6n平方+2)sin1/3n平方+1
用数列极限证明lim(n→∞)(n^-2)/(n^+n+1)=1中证明如下:lim(n→∞)3n+1/5n-4
数列极限(已知lim[(2n-1)an]=2,求lim n*an)
lim(3n^2-2n+6)/(5n^2+7n+9)求数列的极限
数列求极限的问题数列求极限:Xn=(2^n -1)/3^n (n是自然数),那么lim n→∞ Xn=lim n→∞[(2^n -1)/3^n]=多少?
数列极限的运算lim an/(an+1) =2 求lim 2anlim (2n+1)*an=3 求lim n*an
求证一列高数数列极限题:lim(3n^2+n)/(2n^2-1)=3/2
简单的数列极限计算题:lim(3n^2+4n-2)/(2n+1)^2,