Lim (căn4n^2+n-1-cănbac3 (8n^3+n^2+n-1)=0 Moi nguoi xem giup e kq bang 0 dung ko Thank moi nguoi

$\lim (\sqrt{4n^2+n+1}-\sqrt[3]{8n^3+n^2+n-1})$

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

$=\lim\Big( \dfrac{n+1}{\sqrt{4n^2+n+1}+2n} +\dfrac{-n+1}{4n^2+2n.\sqrt[3]{8n^3+n^2+n-1}+\sqrt[3]{8n^3+n^2+n-1}^2}\Big)$

$=\lim \Big( \dfrac{1+\dfrac{1}{n}}{\sqrt{4+\dfrac{1}{n}+\dfrac{1}{n^2}}+2} -\dfrac{ \dfrac{1}{n}-\dfrac{1}{n^2}}{ 4+2\sqrt[3]{8+\dfrac{1}{n}+\dfrac{1}{n^2}-\dfrac{1}{n^3}} +\sqrt[3]{8+\dfrac{1}{n}+\dfrac{1}{n^2}-\dfrac{1}{n^3}}}\Big)$

$=\dfrac{1}{\sqrt4+2}-0$

$=\dfrac{1}{4}$