\(lim\left(2n-\sqrt{8n^3+8n^2+2}\right)=lim\left(2n-2n^{\frac{3}{2}}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\)
\(=lim\left(n\left(1-2\sqrt{n}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\right)=\infty\times\left(-\infty\right)=-\infty\)
lim (2n - \(\sqrt[3]{8n^{3^{ }}+8n^{2^{ }}+2}\) )
\(lim\left(\sqrt{4n^2+2n+1}-\sqrt[3]{8n^3-3n^2+1}\right)\)
Tìm giới hạn lim un
a. \(u_n=\left(2-3n\right)^4\left(n+1\right)^3\)
b.\(u_n=\sqrt[3]{n+4}-\sqrt[3]{n+1}\)
c.\(u_n=\sqrt[3]{8n^3+3n^2+4}-2n+6\)
d. \(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\)
Help me ! Gợi ý cho mik cx đc ạ . Tks mng
lim n(\(\sqrt[3]{n^3-3n^2}-3n\))
lim (\(\sqrt{4n^2+n}+\sqrt[3]{2n^2-8n^3}\))
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 \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
\(1.lim\left(\sqrt[3]{8n^3+4n^2+1}-\sqrt[3]{8n^3-2}\right)\)
\(2.lim\left(\sqrt[3]{n^3+n^2+1}+\sqrt[3]{8-n^3}\right)\)
\(3.lim\left(\sqrt[3]{n^3+n^2+2}-n\right)\)
giá trị của M = lim [(căn bậc ba 1 - n^2 - 8n^3) + 2n] =
Tính: Lim\(\left(\sqrt{n^2+2}.\sqrt[3]{8n^3+1}-\sqrt{4n^2+1}.\sqrt[3]{n^3+2}\right)\)
Tính N = \(lim\left(\sqrt{4n^2+1}-\sqrt[3]{8n^3+n}\right)\)
