Question

lim n(\(\sqrt[3]{n^3-3n^2}-3n\))

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

Answer 1

1) = lim n. \(\frac{n^3-3n^2-27n^3}{\sqrt[3]{\left(n^3-3n^2\right)^2}+3n\sqrt[3]{n^3-3n^2}+9n^2}\)

= lim \(\frac{n\left(-26n^3-3n^2\right)}{\sqrt[3]{\left(n^3-3n^2\right)^2}+3n\sqrt[3]{n^3-3n^2}+9n^2}\)

= lim \(\frac{n^2\left(-26-\frac{3}{n}\right)}{\sqrt[3]{\left(1-\frac{3}{n}\right)^2}+3\sqrt[3]{1-\frac{3}{n}}+9}\)

= lim \(\frac{n^2\left(-26\right)}{13}=-\infty\)

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

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

= \(\frac{1}{2+2}+\frac{2}{4+4+4}=\frac{5}{12}\)