Đáp án:
\[\lim \left( {\sqrt[3]{{8{n^2} + {n^2}}} - \sqrt {4{n^2} - n} } \right) = - \infty \]
Giải thích các bước giải:
\(\begin{array}{l}
\lim \left( {\sqrt[3]{{8{n^2} + {n^2}}} - \sqrt {4{n^2} - n} } \right)\\
= \lim \left( {\sqrt[3]{{9{n^2}}} - \sqrt {4{n^2} - n} } \right)\\
= \lim \left( {\sqrt[3]{{{n^3}.\frac{9}{n}}} - \sqrt {{n^2}\left( {4 - \frac{1}{n}} \right)} } \right)\\
= \lim \left( {n.\sqrt[3]{{\frac{9}{n}}} - n.\sqrt {4 - \frac{1}{n}} } \right)\\
= \lim \left[ {n.\left( {\sqrt[3]{{\frac{9}{n}}} - \sqrt {4 - \frac{1}{n}} } \right)} \right]\\
\lim n = + \infty \\
\lim \left( {\sqrt[3]{{\frac{9}{n}}} - \sqrt {4 - \frac{1}{n}} } \right) = 0 - \sqrt 4 = - 2\\
\Rightarrow \lim \left[ {n.\left( {\sqrt[3]{{\frac{9}{n}}} - \sqrt {4 - \frac{1}{n}} } \right)} \right] = - \infty
\end{array}\)
Vậy \(\lim \left( {\sqrt[3]{{8{n^2} + {n^2}}} - \sqrt {4{n^2} - n} } \right) = - \infty \)
