Đáp án:
\[\lim \left( {\sqrt[3]{{n - {n^3}}} + \sqrt {{n^2} + 3n} } \right) = \frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \left( {\sqrt[3]{{n - {n^3}}} + \sqrt {{n^2} + 3n} } \right)\\
= \lim \left[ {\left( {\sqrt[3]{{n - {n^3}}} + n} \right) + \left( {\sqrt {{n^2} + 3n} - n} \right)} \right]\\
= \lim \left[ {\frac{{n - {n^3} + {n^3}}}{{\sqrt[3]{{{{\left( {n - {n^3}} \right)}^2}}} + n.\sqrt[3]{{n - {n^3}}} + {n^2}}} + \frac{{{n^2} + 3n - {n^2}}}{{\sqrt {{n^2} + 3n} + n}}} \right]\\
= \lim \left[ {\frac{n}{{\sqrt[3]{{{{\left( {n - {n^3}} \right)}^2}}} + n.\sqrt[3]{{n - {n^3}}} + {n^2}}} + \frac{{3n}}{{\sqrt {{n^2} + 3n} + n}}} \right]\\
= \lim \left[ {\frac{{\frac{1}{n}}}{{\sqrt[3]{{{{\left( {\frac{1}{{{n^2}}} - 1} \right)}^2}}} + 1.\sqrt[3]{{\frac{1}{{{n^2}}} - 1}} + 1}} + \frac{3}{{\sqrt {1 + \frac{3}{n}} + 1}}} \right]\\
= \frac{0}{{{{\sqrt[3]{{ - 1}}}^2} + 1.\sqrt[3]{{ - 1}} + 1}} + \frac{3}{{\sqrt 1 + 1}}\\
= \frac{3}{2}
\end{array}\)
