Cho dãy số \(a_n=\sqrt[3]{n^3+1}-n\). Hãy tính \(\lim\limits a_n\).
\(0\).\(\dfrac{1}{3}\).\(\dfrac{1}{2}\).\(1\).Hướng dẫn giải:Có
\(\sqrt[3]{n^3+1}-n=\dfrac{\left(n^3+1\right)-n^3}{\left(\sqrt[3]{n^3+1}\right)^2+n\sqrt[3]{n^3+1}+n^2}\)
\(=\dfrac{1}{n^2\left[\left(\sqrt[3]{1+\dfrac{1}{n^3}}\right)^2+\sqrt[3]{1+\dfrac{1}{n^3}}+1\right]}\)\(=\dfrac{1}{n^2}.\dfrac{1}{\left(\sqrt[3]{1+\dfrac{1}{n^3}}\right)^2+\sqrt[3]{1+\dfrac{1}{n^3}}+1}\)
Suy ra giới hạn cần tính bằng \(0.\)
