Tính limn(căn bậc hai 4n^2+3 - căn bậc ba 8n^3+n) A. dương vô cùng B.1 C.âm vô cùng D.2/3

* Hướng dẫn giải

Ta có: $\lim n\left(\sqrt{4n^2+3}-\sqrt[3]{8n^3+n}\right)$

$=\lim n\left[\left(\sqrt{4n^2+3}-2n\right)+\left(2n-\sqrt[3]{8n^3+n}\right)\right]$

$=\lim \left[n\left(\sqrt{4n^2+3}-2n\right)+n\left(2n-\sqrt[3]{8n^3+n}\right)\right]$

Ta có: $\lim n\left(\sqrt{4n^2+3}-2n\right)=\lim \frac{3n}{\left(\sqrt{4n^2+3}+2n\right)}=\lim \frac{3}{\left(\sqrt{4+\frac{3}{n^2}}+2\right)}=\frac{3}{4}$

Ta có: $\lim n\left(2n-\sqrt[3]{8n^3+n}\right)=\lim \frac{-n^2}{\left(4n^2+2n\sqrt[3]{8n^3+n}+\sqrt[3]{{\left(8n^3+n\right)}^2}\right)}$

$=\lim \frac{-1}{\left(4+2\sqrt[3]{8+\frac{1}{n^2}}+\sqrt[3]{{\left(8+\frac{1}{n^2}\right)}^2}\right)}=-\frac{1}{12}$

Vậy $\lim n\left(\sqrt{4n^2+3}-\sqrt[3]{8n^3+n}\right)=\frac{3}{4}-\frac{1}{12}=\frac{2}{3}$