已知lim(an^2 bn-100) (3n-1)=2,求a,b

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)

3An+Bn=(3An+4Bn)/3+(6An-Bn)/3=>lim(3An+Bn)=lim(3An+4Bn)/3+lim(6An-Bn)/3=3

n→∞则lim[5n-√(an^2+bn+c)]/n=lim2/n=0则lim5-√(an^2+bn+c)/n]=0则√a=5,a=252=lim(5n-√（25n^2+bn+c)）{做分子有理化}=

lim(n->inf)[2n+(an²+2n+1)/(bn+1)]=1lim(2bn²+an²+4n+1)/(bn+1)=1lim[(2b+a)n²+4n+1]

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

n=1-an,第二个式子代入bn=1-anbn+1=(1-an)/(1-an^2)=1/(1+an)an+1=1-bn+1=an/(1+an)求倒数1/(an+1)=1+1/an令cn=1/an,cn

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+

lim(2bn^2+4n+an^2-2n+1)/(bn+2)=1,括号里分子分母同时除以n:lim(2bn+4+an-2+(1/n)/(b+2/n))=1当n趋于无穷时,1/n=2/n=0;要是方程成

lim5an+lim4bn=7lim7an-lim2bn=55liman+4limbn=77liman-2limbn=5liman=17/19limbn=12/19lim(6an+bn)=6

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∴

等于3这是设的,x,y是要求的量再问：求出x和y之后要怎么做？麻烦写一下再答：求出来之后，比如x=1，y=2那么直接带进去，要求的极限值=1×8+2×1=10当然这是我随便举的例子。真正的答案不是这个

设{An}的公差为d1,{Bn}的公差为d2因为limAn/Bn=lim[a1+(n-1)d1]/[b1+(n-1)d2]=lim[a1/n+(1-1/n)d1]/[b1/n+(1-1/n)d2]=(

设{An}的公差为d1,{Bn}的公差为d2因为limAn/Bn=lim[a1+(n-1)d1]/[b1+(n-1)d2]=lim[a1/n+(1-1/n)d1]/[b1/n+(1-1/n)d2]=(

设{an}公差为d,{bn}公差为d'lim(an/bn)=lim[(a1+(n-1)d]/[b1+(n-1)d']=lim[(a1-d)+nd]/[(b1-d')+nd']=lim[(a1-d)/n

令x（2an+4bn）+y（6an-bn）=3an+bn,则2x+6y=3,4x-y=1,易解得x=9/26,y=5/13,所以lim（3an+bn）=72/26+5/13=41/13

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

已知：lim[√(n^2+a*n)-(b*n+1)]=b,求a.因为√(n^2+a*n)-(b*n+1)=[√(n^2+a*n)^2-(b*n+1)^2]/[√(n^2+a*n)+(b*n+1)]（分

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

a+b=-4