limn+无穷
@夹宇1582：limn~无穷(1+x)(1+x2)(1+x4)..,(1+x2n) - 作业帮
漆琦14787436853…… [答案] (1+x)+.+(1+xn) =x^(n+1)*n/2=x^(n*n/2)
@夹宇1582：limn→ 无穷=you什么意思? -
漆琦14787436853…… 两种解释都解释得通,一种是“永远都是你!”当n趋于无穷时,就是你,那么永远都是你,前提是现在就是你.另一种解释是,你排在最后,除非到了n无穷时,才轮到你,也就是相当于说”世界上的人都死绝了,也不一定轮到你“.前提是现在不是你.
@夹宇1582：证明limn→无穷[(1/n2+n+1)+(2/n2+n+2)+……+(n/n2+n+n)]=½ - 作业帮
漆琦14787436853…… [答案] 当n→无穷, 1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n) ≤1/(n^2+n+1)+2/(n^2+n+1)+...+n/(n^2+n+1) =(1+2+...+n)/(n^2+n+1) =(1/2)(n+1)n/(n^2+n+1) -> 1/2 1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n) ≥1/(n^2+n+n)+2/(n^2+n+n)+...+n/(n^2+n+n) = (1+2+...+...
@夹宇1582：利用极限的夹副准则证明limn→无穷大(n/n^2+π+n/n^2+2π+...+n/n^2+nπ)=1 - 作业帮
漆琦14787436853…… [答案] n/(n^2+nπ) ≤ n/(n^2+mπ) ≤ n/(n^2 + π) 注:n ≤ m ≤ 1 所以, n*[n/(n^2+nπ)]=n^2/(n^2+nπ) ≤ ∑n/(n^2+mπ) ≤ n*[n/(n^2+π) = n^2/(n^2+π) 因为:lim[n^2/(n^2+nπ)]=lim[1/(1+π/n)] = 1 lim[n^2/(n^2+π)] = lim[1/(1+π/n^2)] = 1 所以, lim∑n/(n^2+mπ) = 1
@夹宇1582：limn趋近于无穷2·5^n+3^n/5^n+1+2^n+1 - 作业帮
漆琦14787436853…… [答案] lim(n→∞)(2·5^n+3^n)/[5^(n+1)+2^(n+1)] =lim(n→∞)[2+(3/5)^n]/[5+2*(2/5)^n] =(2+0)/(5+0) =2/5 如果不懂,祝学习愉快!
@夹宇1582：limn趋于无穷(√n +√2n +……+ √(n - 1)n +√n^2)/n^2 - 作业帮
漆琦14787436853…… [答案] ≧(√n+√n+.+√n)/n^2=1/√n=0
@夹宇1582：limn–无穷 n[In(n+1) - In(n - 1)] - 作业帮
漆琦14787436853…… [答案] limn–无穷 n[In(n+1)-In(n-1)]=limn–无穷 [In(n+1)/(n-1)]/(1/n)=limn–无穷 [In(n-1+2)/(n-1)]/(1/n)=limn–无穷 [In(1+2/(n-1))/(1/n)]=limn–无穷 [2/(n-1))/(1/n)]=limn–无穷 [2n/(n-1))]=limn–无穷 [2/(1-1...
@夹宇1582：limn→正无穷 n[(1+1/n)∧n - e] - 作业帮
漆琦14787436853…… [答案] Taylor展式:ln(1+1/n)=1/n-1/(2n^2)+小o(1/n^2),e^(-1/(2n)+小o(1/n))=1-1/(2n)+小o(1/n). lim n[(1+1/n)^n-e]=lim n[e^(nln(1+1/n))-e]=lim n[e^(1-1/(2n)+小o(1/n))-e]=lim en[e^(-1/(2n)+小o(1/n))-1]=lim ne[-1/(2n)+小o(1/n)]=-e/2.
@夹宇1582：Limn趋于无穷〔根号(n+2) - 根号(n+1)+根号n] - 作业帮
漆琦14787436853…… [答案] lim[√(n+2) - √(n+1) + √n] n→∞ ∵lim[√(n+2) - √(n+1)] n→∞ = lim[√(n+2)-√(n+1)][√(n+2)+√(n+1)]/[√(n+2)+√(n+1)] [分子有理化] n→∞ = lim 1/[√(n+2)+√(n+1)] n→∞ = 1/[√∞+√∞] = 1/∞ = 0 ∴lim[√(n+2) - √(n+1) + √n] n→∞ = lim[0 + √n] n→∞ = ∞
@夹宇1582：limn趋近于无穷[1+2+3+...+(n+1)]/n∧2 - 作业帮
漆琦14787436853…… [答案] =limn趋近于无穷[(n+1)(n+2)/2]/n^2 =limn趋近于无穷(n^2+3n+2)/[2n^2] 分子分母同除以n^2 所以最后极限就=1/2
漆琦14787436853…… [答案] (1+x)+.+(1+xn) =x^(n+1)*n/2=x^(n*n/2)
@夹宇1582：limn→ 无穷=you什么意思? -
漆琦14787436853…… 两种解释都解释得通,一种是“永远都是你!”当n趋于无穷时,就是你,那么永远都是你,前提是现在就是你.另一种解释是,你排在最后,除非到了n无穷时,才轮到你,也就是相当于说”世界上的人都死绝了,也不一定轮到你“.前提是现在不是你.
@夹宇1582：证明limn→无穷[(1/n2+n+1)+(2/n2+n+2)+……+(n/n2+n+n)]=½ - 作业帮
漆琦14787436853…… [答案] 当n→无穷, 1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n) ≤1/(n^2+n+1)+2/(n^2+n+1)+...+n/(n^2+n+1) =(1+2+...+n)/(n^2+n+1) =(1/2)(n+1)n/(n^2+n+1) -> 1/2 1/(n^2+n+1)+2/(n^2+n+2)+...+n/(n^2+n+n) ≥1/(n^2+n+n)+2/(n^2+n+n)+...+n/(n^2+n+n) = (1+2+...+...
@夹宇1582：利用极限的夹副准则证明limn→无穷大(n/n^2+π+n/n^2+2π+...+n/n^2+nπ)=1 - 作业帮
漆琦14787436853…… [答案] n/(n^2+nπ) ≤ n/(n^2+mπ) ≤ n/(n^2 + π) 注:n ≤ m ≤ 1 所以, n*[n/(n^2+nπ)]=n^2/(n^2+nπ) ≤ ∑n/(n^2+mπ) ≤ n*[n/(n^2+π) = n^2/(n^2+π) 因为:lim[n^2/(n^2+nπ)]=lim[1/(1+π/n)] = 1 lim[n^2/(n^2+π)] = lim[1/(1+π/n^2)] = 1 所以, lim∑n/(n^2+mπ) = 1
@夹宇1582：limn趋近于无穷2·5^n+3^n/5^n+1+2^n+1 - 作业帮
漆琦14787436853…… [答案] lim(n→∞)(2·5^n+3^n)/[5^(n+1)+2^(n+1)] =lim(n→∞)[2+(3/5)^n]/[5+2*(2/5)^n] =(2+0)/(5+0) =2/5 如果不懂,祝学习愉快!
@夹宇1582：limn趋于无穷(√n +√2n +……+ √(n - 1)n +√n^2)/n^2 - 作业帮
漆琦14787436853…… [答案] ≧(√n+√n+.+√n)/n^2=1/√n=0
@夹宇1582：limn–无穷 n[In(n+1) - In(n - 1)] - 作业帮
漆琦14787436853…… [答案] limn–无穷 n[In(n+1)-In(n-1)]=limn–无穷 [In(n+1)/(n-1)]/(1/n)=limn–无穷 [In(n-1+2)/(n-1)]/(1/n)=limn–无穷 [In(1+2/(n-1))/(1/n)]=limn–无穷 [2/(n-1))/(1/n)]=limn–无穷 [2n/(n-1))]=limn–无穷 [2/(1-1...
@夹宇1582：limn→正无穷 n[(1+1/n)∧n - e] - 作业帮
漆琦14787436853…… [答案] Taylor展式:ln(1+1/n)=1/n-1/(2n^2)+小o(1/n^2),e^(-1/(2n)+小o(1/n))=1-1/(2n)+小o(1/n). lim n[(1+1/n)^n-e]=lim n[e^(nln(1+1/n))-e]=lim n[e^(1-1/(2n)+小o(1/n))-e]=lim en[e^(-1/(2n)+小o(1/n))-1]=lim ne[-1/(2n)+小o(1/n)]=-e/2.
@夹宇1582：Limn趋于无穷〔根号(n+2) - 根号(n+1)+根号n] - 作业帮
漆琦14787436853…… [答案] lim[√(n+2) - √(n+1) + √n] n→∞ ∵lim[√(n+2) - √(n+1)] n→∞ = lim[√(n+2)-√(n+1)][√(n+2)+√(n+1)]/[√(n+2)+√(n+1)] [分子有理化] n→∞ = lim 1/[√(n+2)+√(n+1)] n→∞ = 1/[√∞+√∞] = 1/∞ = 0 ∴lim[√(n+2) - √(n+1) + √n] n→∞ = lim[0 + √n] n→∞ = ∞
@夹宇1582：limn趋近于无穷[1+2+3+...+(n+1)]/n∧2 - 作业帮
漆琦14787436853…… [答案] =limn趋近于无穷[(n+1)(n+2)/2]/n^2 =limn趋近于无穷(n^2+3n+2)/[2n^2] 分子分母同除以n^2 所以最后极限就=1/2
