Cho biết giá trị của \(\lim (\sqrt {{n^2} + 2n}  - \sqrt[3]{{{n^3} + 2{n^2}}})\) bằng ?

* Hướng dẫn giải

\(\begin{array}{l}\lim \left( {\sqrt {{n^2} + 2n}  - \sqrt[3]{{{n^3} + 2{n^2}}}} \right)\\ = \lim \left( {\sqrt {{n^2} + 2n}  - n} \right) \\+ \lim \left( {n - \sqrt[3]{{{n^3} + 2{n^2}}}} \right)\\ = \lim \dfrac{{{n^2} + 2n - {n^2}}}{{\sqrt {{n^2} + 2n}  + n}} \\+ \lim \dfrac{{{n^3} - {n^3} - 2{n^2}}}{{{n^2} + n.\sqrt[3]{{{n^3} + 2{n^2}}} + {{\left( {\sqrt[3]{{{n^3} + 2{n^2}}}} \right)}^2}}}\\ = \lim \dfrac{{2n}}{{\sqrt {{n^2} + 2n}  + n}} \\+ \lim \dfrac{{ - 2{n^2}}}{{{n^2} + n.\sqrt[3]{{{n^3} + 2{n^2}}} + {{\left( {\sqrt[3]{{{n^3} + 2{n^2}}}} \right)}^2}}}\\ = \lim \dfrac{2}{{\sqrt {1 + \dfrac{2}{n}}  + 1}} \\+ \lim \dfrac{{ - 2}}{{1 + \sqrt[3]{{1 + \dfrac{2}{n}}} + {{\left( {\sqrt[3]{{1 + \dfrac{2}{n}}}} \right)}^2}}}\\ = 1 + \left( { - \dfrac{2}{3}} \right) = \dfrac{1}{3}\end{array}\)