证明n趋向无穷时,5n^2/(7n-n^2)的极限等于-5
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/27 22:10:54
x){ٌۺNl+?M4/H_l^˙|O&HLv6ؚ
,dGד@z^\UjjB@a\ 1L0i{=꘧ 7hCv˦OvvHyD0$a.0 OG
证明n趋向无穷时,5n^2/(7n-n^2)的极限等于-5
证明n趋向无穷时,5n^2/(7n-n^2)的极限等于-5
证明n趋向无穷时,5n^2/(7n-n^2)的极限等于-5
lim5n^2/(7n-n^2)
上下同除n^2
=lim(5/(7/n-1))
=5/(lim7/n-1)
=5/(0-1)
=-5
原式=lim(n->∞) 5/(7/n-1)
=5/(0-1)
=-5
lim5n^2/(7n-n^2)
上下同除n^2
那么
原式=lim(5/(7/n-1))
=5/(lim7/n-1)
=5/(0-1)
=-5
证明n趋向无穷时,5n^2/(7n-n^2)的极限等于-5
证明当n趋向于无穷时数列an=(1*3*5*7*...*2n-1)/(2*4*6*8*...*2n)趋向于0?
lim(2^n+3^n)^1
(n趋向无穷)
(2^n*n!)/n^n在n趋向无穷时等于多少?
n趋向于无穷
证明数列sin n(n为正整数)当n趋向正无穷时极限不存在
用极限定义证明: lim( 2^n/n!)=0 其中n趋向于无穷.
依据极限定义证明lim{(n^2+a^2)/n}=1 n趋向于无穷时
当n趋向于无穷时,求n^(1/2)*[n^(1/n)-1]的极限
当n趋向无穷时,1/(n^2+1)+2/(n^2+2)+.+n/(n^2+n)的极限是多少
在n趋向无穷时(n+1分之一+n+2分之2+.+n+n分之n)的极限
求n趋向无穷时 [(1+1/n)(1+2/n)...(1+n/n)]^1/n 的极限?
计算数列极限,当N趋向于无穷时,根号下(N^2+4N+5)-(N-1)的极限
(2n^3+n^2-6n+7)/(3n^3+4n^2-1),n趋向于无穷时的极限值是多少
为什么lim n趋向无穷 (5n^2-4n)/(3n^2+2n-1)等于5/3
lim[(-2)^n+3^n]/[(-2)^(n+1)+3^(n+1)] 当n趋向于无穷时
求极限 n趋向无穷 2^n+1 + 3^n+1/2^n+3^n
lim(n/(n^2+1)+...+n/(n^2+n))x趋向于无穷 求解答过程~