\(=lim\left(\frac{1}{2}+\frac{\left(-\sqrt{3}\right)^n}{6.2^n}\right)=lim\left(\frac{1}{2}+\frac{\left(\frac{-\sqrt{3}}{2}\right)^n}{6}\right)=\frac{1}{2}+\frac{0}{6}=\frac{1}{2}\)
lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
lim \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\)
lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)
a)lim \(\frac{\left(2n+1\right)^2\left(n-1\right)}{\sqrt[3]{n^3+7n-2}}\)
b)lim [(2n-1)\(\sqrt{\frac{2n^2+5}{n^4+n^2+2}}\)]
c)lim [n(\(\sqrt[3]{n^3+n^2}-n\))]
\(lim\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+......+\frac{1}{\sqrt{n^2+n}}\right)\)
Tìm giới hạn dãy số sau
\(lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}\)
\(lim\left(3.2^{n+1}-5.3^n+7n\right)\)
1
a,Lim\(\sqrt{1+2n-n^3}\)
b,Lim\(\sqrt{n^2+2n+3}-\sqrt[3]{n^2+n^3}\)
c,Lim\(\dfrac{\left(2\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n+1\right)\left(n+2\right)}\)
d,\(\dfrac{4^{n+1}-3\times2^n}{3^{n+2}+2^n}\)
e,\(\dfrac{7^{n+1}-5^{n+2}+3}{2\times6^{n+1}-3^n+3}\)
f,\(\dfrac{\sqrt{n^4+1}}{n}\) -\(\dfrac{\sqrt{4n^6+1}}{n}\)
\(1.lim\left(\sqrt[3]{8n^3+4n^2+1}-\sqrt[3]{8n^3-2}\right)\)
\(2.lim\left(\sqrt[3]{n^3+n^2+1}+\sqrt[3]{8-n^3}\right)\)
\(3.lim\left(\sqrt[3]{n^3+n^2+2}-n\right)\)
cho f(n) = \(\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\frac{1}{\sqrt[3]{4}}+...+\frac{1}{\sqrt[3]{n}}\) nϵN*. GIá trị lim\(\frac{f\left(n\right)}{n^2+1}\) bằng ?
