数列极限的题目.求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))
来源:学生作业帮助网 编辑:六六作业网 时间:2025/02/07 19:23:27
数列极限的题目.求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))数列极限的题目.求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))数列极限的题目.
数列极限的题目.求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))
数列极限的题目.
求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))
数列极限的题目.求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))
lim{[√(2n+3)-√(2n-1)]/[√(3n+9)-√(3n+16)]}
=lim{[√(2n+3)-√(2n-1)][√(3n+9)+√(3n+16)]/[(3n+9)-(3n+16)]}
=-(1/7)lim{[√(2n+3)-√(2n-1)][√(3n+9)+√(3n+16)]}
=-(1/7)lim{[√(3n+9)+√(3n+16)]/{1/[√(2n+3)-√(2n-1)]}
=-(1/7)lim{{[√(3n+9)+√(3n+16)]/{[√(2n+3)+√(2n-1)]/[(2n+3)-(2n-1)]}}
=-(4/7)lim{[√(3n+9)+√(3n+16)]/[√(2n+3)+√(2n-1)]}
=-(4/7)lim{[√(3+9/n)+√(3+16/n)]/[√(2+3/n)+√(2-1/n)]}
=-(4/7){[√3+√3]/[√2+√2]}
=-(4/7)[√3/√2]
=-(2/7)√6
数列极限的题目.求lim(√(2n+3)-√(2n-1)/√(3n+9)-√(3n+16))
求下列数列的极限:lim (2+3^n)/(1+3^(n+1))
lim n →∞ (1^n+3^n+2^n)^1/n,求数列极限
lim(3n^2-2n+6)/(5n^2+7n+9)求数列的极限
数列极限的运算lim an/(an+1) =2 求lim 2anlim (2n+1)*an=3 求lim n*an
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?
一道数列极限的题目!lim(2n-根号下(4n2+kn+3))=1求k取值范围
数列极限题目.求lim(cosa^n+sina^n)/(cos^n-sina^n)(0
高数 数列极限lim(1+ 2^n + 3^n)^(1/n) n趋于无穷大求极限
求数列的极限:lim(2^n)/[1+2^(n+1)]
数列极限的题目已知lim(n趋向无穷大)(5n-根号(an^2-bn+c))=2,求a,b的值
求数列极限lim(6n平方+2)sin1/3n平方+1
数列的极限计算lim(3n²+4n-2)/(2n+1)²
数列极限(已知lim[(2n-1)an]=2,求lim n*an)
两个数列求An/Bn极限我已经求出两个数列An=2^(n+3)-14*3^(n-1)以及Bn=28*3^(n-1)-3*2^(n+2)题目让求lim(An/Bn)
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