lim(n→∞)(2n-1/2n+1)^n= 大一高数~
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/16 21:39:46
lim(n→∞)(2n-1/2n+1)^n=大一高数~lim(n→∞)(2n-1/2n+1)^n=大一高数~lim(n→∞)(2n-1/2n+1)^n=大一高数~lim(n→∞)(2n-1/2n+1)
lim(n→∞)(2n-1/2n+1)^n= 大一高数~
lim(n→∞)(2n-1/2n+1)^n= 大一高数~
lim(n→∞)(2n-1/2n+1)^n= 大一高数~
lim(n→∞)(2n-1/2n+1)^n=lim(n→∞)[2n+1/2n-1)]^(-n)=lim(n→∞)[1+(1/2n-1)]^(-n)
=1/e
lim (n!+(n-1)!+(n-2)!+(N-3)!+⋯..+2!+1)/n!其中n→∞
lim(n→∞) ((2n!/n!*n)^1/n的极限用定积分求是lim(n→∞) 1/n(2n!/n!)^1/n 不好意思
求lim n→∞ (1+2/n)^n+3
lim(n→∞)[1-(2n/n+3)]
lim(n→∞)(2n-1/n+3)
lim(n→∞)[1/(3n+1)+1/(3n+2)+~1/(3n+n)]
求极限lim [ 2^(n+1)+3^(n+1)]/2^n+3^n (n→∞)
lim(n→∞) 根号n+2-根号n+1/根号n+1-根号n
计算lim(n→∞)(1^n+2^n+3^n)^(1/n)
lim n →∞ (1^n+3^n+2^n)^1/n,求数列极限
lim(n→∞)(3n^3-2n+1)/n^3+n^2 快
求极限lim(x→∞)(1/n+2/n+3/n..+n/n)
用夹逼定理求lim(n→∞)[√(n^2+n)-n]^(1/n)
求极限lim(n→∞)(1/n²+2/n²+...+n/n²)
用夹逼定理求lim(n→∞)√[(n^2+n)-n]^(1/n)
求lim n→+∞(1/n^k+2/n^k+ +n/n^k)有三种情况,
lim x→n (√n+1-√n)*√(n+1/2)lim x n→∞ (√n+1-√n)*√(n+1/2)
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)