Question

giá trị của \(K=lim\left(\sqrt[3]{n^3+n^2-1}-3\sqrt{4n^2+n+1}+5n\right)\) bằng :

A. \(-\frac{5}{12}\)

B. \(-\infty\)

C. 1

D. \(+\infty\)

Answer 1

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

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

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

\(=\frac{1}{1+1+1}-\frac{3}{2+2}=-\frac{5}{12}\)