已知lim(an² 5n-2 3n 1 -n)

lim(5n)-lim((an^2-bn+c)^0.5)=2lim(5n-2)=lim((an^2-bn+c)^0.5)根据lim的唯一性,可知5n-2=(an^2-bn+c)^0.5即：(5n-2)

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因为liman=A,所以{an}有界,不妨设为M且对任意e>0,有存在N,当n>N时|an-A|0|an^2-A^2|=|an+A||an-A|N时|an^2-A^2|

lim[(2n-1)an]=lim{[(2n-1)/n]*n*an}因为llim(2n-1)/n=2所以lim[(2n-1)an]=2lim(n*an)=1推知：lim(n*an)=1/2

liman＝lim[(2n+1)an]/(2n+1)＝lim[(2n+1)an]×lim1/(2n+1)＝3×0＝0所以,3＝lim[(2n+1)an]＝2×limnan＋liman＝2×limnan

lim(n^2+1/n+1+an-b)=lim(n^2+1/n+1+(an-b)(n+1)/n+1)=lim[(1+a)n^2+(a-b)n+(1-b)]/(n+1)又lim(n^2+1/n+1+an

lim{[(3n^2+cn+1)/(an^2+bn)]-4n}=5lim{[(3n^2+cn+1)-4n(an^2+bn)]/(an^2+bn)}=5lim{[-4an^3+(3-4b)n^2+cn+

其实这个结论不需要什么过程的吧.一定要写过程的话就是lim(3n-2)an=6所以lim(3n*an)=6所以3lim(n*an)=6所以就是2

lim[(2n-1)an]=2lim[(2n-1)an]/[(n+2)an]=lim(2n-1)/(n+2)=2=>lim(n+2)an=1

首先a＝0,否则极限不存在.又lim(n→∞)[(an^2+bn+c)/(2n+5)]＝lim(n→∞)[(bn+c)/(2n+5)]＝lim(n→∞)[(b+c/n)/(2+5/n)]＝b/2=3∴

楼上两位可能将题意理解错了.第一题题意不明,有两种可能.详解见图.点击放大,再点击再放大.

因为iman=0(n->无穷)所以对于任意小的e>0都存在N,使得当n>N时使得an

是填空还是解答题?填空可以用赋值法,令an=2n,bn=n,马上得出答案1/2设an=a1+(n-1)d1bn=b1+(n-1)d2,其中d1,d2均不为0lim(n趋近无穷)an/bn=2得d1=2

lim(n→∞)[(n^2+n)/(n+1)-an-b]=lim(n→∞)[(1-a)n-b]=1,显然a=1,使(1-a)n=0.-b=1,所以b=-1.

n趋于无穷大看最高项,因此lim(n趋于无穷大)(5n^2+3n)/(an^2+b)＝lim(n趋于无穷大)5n^2/an^2＝5/a

lim[(3n^2+cn+1)/(an^2+bn)-4n]=lim[(3n^2+cn+1-4an^3-4bn^2)/(an^2+bn)]则-4a=0即a=0极限化成lim[(3n^2+cn+1-4bn

很显然,如果a不为0,该极限不存在,因为an^2+bn+5是3n-2的高阶无穷大所以a=0（bn+5)/(3n-2)=(b+5/n)(3-2/n)5/n,2/n可以忽略,所以极限等于b/3=2,所以b

n=k,a1/1^2+a2/2^2+a3/3^2.+ak/k^2＞3^kn=k+1,a1/1^2+a2/2^2+a3/3^2.+ak/k^2+a(k+1)/(k+1)^2>3^k+a(k+1)/(k+

对于数列{an},已知limn→∞n*an=5,求limn→∞(3n+7)an的值5-离问题结束还有14天23小时1.对于数列{an},已知limn→∞n*an=5,求limn→∞(3n+7)an的值