Cho dãy số \(a_n=\sqrt[3]{n^3-3n^2+1}-\sqrt{n^2+4n}\). Hãy tính \(\lim\limits a_n\) .
\(2\).\(-2\).\(3\).\(-3\).Hướng dẫn giải:Có \(a_n=\sqrt[3]{n^3-3n^2+1}-n-\left(\sqrt{n^2+4n}-n\right)\)
\(u_n=\sqrt[3]{n^3-3n^2+1}-n=\dfrac{\left(n^3-3n^2+1\right)-n^3}{\left(\sqrt[3]{n^3-3n^2+1}\right)^2+n\sqrt[3]{n^3-3n^2+1}+n^2}\)
\(=\dfrac{-3+\dfrac{1}{n^2}}{\left(\sqrt[3]{1-\dfrac{3}{n}+\dfrac{1}{n^3}}\right)^2+\sqrt[3]{1-\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\),
\(v_n=\sqrt{n^2+4n}-n=\dfrac{4n}{\sqrt{n^2+4n}+n}=\dfrac{4}{\sqrt{1+\dfrac{4}{n}}+1}\)
\(\lim u_n=-1,\)\(\lim v_n=2\) suy ra \(\lim a_n=-3.\)
