Tính lim (căn bậc ba (n^3 + 8n^2) - n)
Câu hỏi:
Tính \(\lim \left( {\sqrt[3]{{{n^3} + 8{n^2}}} - n} \right)\).
Trả lời:
\(\lim \left( {\sqrt[3]{{{n^3} + 8{n^2}}} - n} \right)\)
\( = \lim \frac{{\left( {\sqrt[3]{{{n^3} + 8{n^2}}} - n} \right)\left( {\sqrt[3]{{{{\left( {{n^3} + 8{n^2}} \right)}^2}}} + n\sqrt[3]{{{n^3} + 8{n^2}}} + {n^2}} \right)}}{{\sqrt[3]{{{{\left( {{n^3} + 8{n^2}} \right)}^2}}} + n\sqrt[3]{{{n^3} + 8{n^2}}} + {n^2}}}\)
\[ = \lim \frac{{8{n^2}}}{{\sqrt[3]{{{{\left( {{n^3} + 8{n^2}} \right)}^2}}} + n\sqrt[3]{{{n^3} + 8{n^2}}} + {n^2}}}\]
\[ = \lim \frac{8}{{\sqrt[3]{{{{\left( {1 + \frac{8}{n}} \right)}^2}}} + \sqrt[3]{{1 + \frac{8}{n}}} + 1}}\]
\[ = \frac{8}{3}\].
