lim(3^2n+5^n)/(1+9^n)3^2n or 5^n
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/26 03:42:17
lim(3^2n+5^n)/(1+9^n)3^2nor5^nlim(3^2n+5^n)/(1+9^n)3^2nor5^nlim(3^2n+5^n)/(1+9^n)3^2nor5^n除以9^n,3^2n
lim(3^2n+5^n)/(1+9^n)3^2n or 5^n
lim(3^2n+5^n)/(1+9^n)
3^2n or 5^n
lim(3^2n+5^n)/(1+9^n)3^2n or 5^n
除以9^n ,3^2n就是9^n
lim(3^2n+5^n)/(1+9^n) =lim(9^n+5^n)/(1+9^n) =lim[1+(5/9)^n]/(1/9^n+1) =(1+0)/(0+1) =1 这种题目,一般除以比较大的,即9^n
lim(3^2n+5^n)/(1+9^n)=?
lim(3^2n+5^n)/(1+9^n)3^2n or 5^n
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)(4-n)/(n-1)(3-2n)
lim(n^3+n)/(n^4-3n^2+1)
lim[(n+3)/(n+1))]^(n-2) 【n无穷大】
lim(2^n+3^n)^1
(n趋向无穷)
lim (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中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(n->无穷)[(1+3+5+.(2n+1)]/n^2]^(3-2n)
lim((2n+1)^5(n+10)^3)/((n^2-2n-3)^4)=32
lim(n-无穷)=3n^2/5n^2+2n-1,
求lim(n->无穷)(3n+5)/(n^2+n+4)^1/2的极限
lim(n->∞) n+1/n^2 - 9
lim n->无穷大(2^n-1)/(3^n+1)
lim根号n^2+n+1/3n-2
lim根号n^2+n+1/3n-2=?
求lim n→∞ (1+2/n)^n+3