夹逼定理求极限limn[1/(n^2+π)+1/(n^2+2π)……+1/(n^2+nπ)]=1
夹逼定理求极限limn[1/(n^2+π)+1/(n^2+2π)……+1/(n^2+nπ)]=1
用夹逼定理求极限运用夹逼定理求下列序列的极限(6n^4+n-2)^(1/n)(lg3n)^(1/n)[2/(3n^2-n
求极限limn→∞(n-1)^2/(n+1)
求极限:limn→∞(n-1)^2/(n+1)
求下列数列极限(1)limn→∞2n^3-n+1/n^3+2n^2;(2)limn→∞(-2)^n+3^n/(-2)^n
求极限limn趋于无穷 1/n^2+2/n^2+...+n-1/n^2+n/n^2
利用极限存在准则证明:limn趋向于无穷,n【1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)】=
limn→∞,n/(√n^2+1)+(√n^2-1)求极限
求极限limn→∞(n^2)*(k/n -1/(n+1)-1/(n+2)...-1/(n+k)
求limn→∞((3^n+2^n)/(3^(n+1)-2^(n+1)))的极限
夹逼定理求极限 答案是1/2 n趋近于无穷大
计算:limn^2[(k/n)-(1/n+1)-(1/n+2)-……-(1/n+k)]