lim[4-2^(n+1)/2^n+2^(n+2)],n→∝
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/18 22:55:56
![lim[4-2^(n+1)/2^n+2^(n+2)],n→∝](/uploads/image/z/12523699-19-9.jpg?t=lim%5B4-2%5E%28n%2B1%29%2F2%5En%2B2%5E%28n%2B2%29%5D%2Cn%E2%86%92%E2%88%9D)
x)̍65674cuMz1&Hev6D|{f=h{v|鄞3<ٽ^O;6<Ta>TBPqD}
C}m{:u5m:C|Vb>DTDYgК'?7ٌ E@)M[]#}S=<;l<{@(6b!r S
lim[4-2^(n+1)/2^n+2^(n+2)],n→∝
lim[4-2^(n+1)/2^n+2^(n+2)],n→∝
lim[4-2^(n+1)/2^n+2^(n+2)],n→∝
答:分子分母同除以2^(n+2),得到式子【4/2^(n+2)-1/2】/(1/4+1),当n→∝时,分子等于-1/2,分母等于5/4,所以结果是(-1/2)/(5/4)=-2/5.
lim[4-2^(n+1)/2^n+2^(n+2)],
=lim[4-2*2^n/2^n+4*2^n]
=-2/5
lim(n+3)(4-n)/(n-1)(3-2n)
lim(n^3+n)/(n^4-3n^2+1)
lim[(4+7+...+3n+1)/(n^2-n)]=
lim (1+2/n)^n+4 n-->无穷大 求极限
lim(1/n+2/n+3/n+4/n+5/n+……+n/n)=lim(1/n)+lim(2/n)+……+lim(n/n)成立吗?(n趋近于无穷大)为什么不成立?
lim[(n+3)/(n+1))]^(n-2) 【n无穷大】
lim(2^n+3^n)^1
(n趋向无穷)
lim(1-1/n)^(n^2)=?
lim(1-1/n^2)^n=?
lim(1/n^2+4/n^2+7/n^2+…+3n-1/n^2)
求lim(n+1)(n+2)(n+3)/(n^4+n^2+1)
lim n趋向无穷大3n^3+n^2-3/4n^3+2n+1
一道极限题,lim[n^2(2n+1)]/(n^3+n+4)n->∞
1、lim n->无穷 根号[(n^4+n+1)-n^2]*(3n+4)
求极限 lim(n→∞) tan^n (π/4 + 2/n) lim(n→∞)tan^n(π/4+2/n) =lim(n→∞)[(tan(π/4)+tan(2/n))/(1-tan(π/4)tan(2/n))]^n =lim(n→∞)[(1+tan(2/n))/(1-tan(2/n))]^n =lim(n→∞)(1+tan(2/n))^n/(1-tan(2/n))^n (1) 因为 lim(n→∞)(1+tan(2/n)
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
lim In(n+1) / In(n+2) 是多少lim [In(n+1) / In(n+2)] 是多少
求1.lim(3n-(3n^2+2n)/(n-1)) 2.lim(8+1/(n+1)) 3.lim根号n(根号(n+1)-根号(n-3))