giải cho em câu này với ạ lim(cănbậcba(8n^3-n) – 2n)

Đáp án:

\[\lim \left( {\sqrt[3]{{8{n^3} – n}} – 2n} \right) = 0\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\lim \left( {\sqrt[3]{{8{n^3} – n}} – 2n} \right)\\
 = \lim \frac{{\left( {\sqrt[3]{{8{n^3} – n}} – 2n} \right)\left( {{{\sqrt[3]{{8{n^3} – n}}}^2} + 2n.\sqrt[3]{{8{n^3} – n}} + 4{n^2}} \right)}}{{{{\sqrt[3]{{8{n^3} – n}}}^2} + 2n.\sqrt[3]{{8{n^3} – n}} + 4{n^2}}}\\
 = \lim \frac{{{{\left( {\sqrt[3]{{8{n^3} – n}}} \right)}^3} – {{\left( {2n} \right)}^3}}}{{{{\sqrt[3]{{8{n^3} – n}}}^2} + 2n.\sqrt[3]{{8{n^3} – n}} + 4{n^2}}}\\
 = \lim \frac{{ – n}}{{{{\sqrt[3]{{8{n^3} – n}}}^2} + 2n.\sqrt[3]{{8{n^3} – n}} + 4{n^2}}}\\
 = \lim \frac{{ – \frac{1}{n}}}{{{{\sqrt[3]{{8 – \frac{1}{{{n^2}}}}}}^2} + 2.\sqrt[3]{{8 – \frac{1}{{{n^2}}}}} + 4}}\\
 = \frac{0}{{{{\sqrt[3]{8}}^2} + 2.\sqrt[3]{8} + 4}} = 0
\end{array}\)