Giải thích các bước giải:
3.Ta có:
$\lim(\sqrt{4n^2+2n}-2n^2)$
$=\lim n^2(\sqrt{\dfrac4{n^2}+\dfrac2{n^3}}-2)$
$=+\infty(\sqrt{0+0}-2)$
$=-\infty$
4.Ta có:
$\lim(\sqrt[3]{8n^3+2n}-2n)$
$=\lim\dfrac{8n^3+2n-(2n)^3}{(\sqrt[3]{8n^3+2n})^2+2n\sqrt[3]{8n^3+2n}+4n^2}$
$=\lim\dfrac{2n}{(\sqrt[3]{8n^3+2n})^2+2n\sqrt[3]{8n^3+2n}+4n^2}$
$=\lim\dfrac{\dfrac2n}{(\sqrt[3]{8+\dfrac2{n^2}})^2+2\sqrt[3]{8+\dfrac2{n^2}}+4}$
$=\dfrac{0}{(\sqrt[3]{8+0})^2+2\sqrt[3]{8+0}+4}$
$=0$
