Đáp án:
\[\lim \frac{{{n^2} + \sqrt[3]{{1 - {n^6}}}}}{{\sqrt {{n^4} + 1} - {n^2}}} = \frac{2}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\lim \frac{{{n^2} + \sqrt[3]{{1 - {n^6}}}}}{{\sqrt {{n^4} + 1} - {n^2}}}\\
= \lim \left[ {\left( {{n^2} + \sqrt[3]{{1 - {n^6}}}} \right):\left( {\sqrt {{n^4} + 1} - {n^2}} \right)} \right]\\
= \lim \left[ {\frac{{{{\left( {{n^2}} \right)}^3} + {{\sqrt[3]{{1 - {n^6}}}}^3}}}{{{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}:\frac{{{{\sqrt {{n^4} + 1} }^2} - {{\left( {{n^2}} \right)}^2}}}{{\sqrt {{n^4} + 1} + {n^2}}}} \right]\\
= \lim \left[ {\frac{1}{{{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}:\frac{1}{{\sqrt {{n^4} + 1} + {n^2}}}} \right]\\
= \lim \frac{{\sqrt {{n^4} + 1} + {n^2}}}{{{n^4} - {n^2}.\sqrt[3]{{1 - {n^6}}} + {{\sqrt[3]{{1 - {n^6}}}}^2}}}\\
= \lim \frac{{\sqrt {1 + \frac{1}{{{n^4}}}} + 1}}{{1 - 1.\sqrt[3]{{\frac{1}{{{n^6}}} - 1}} + {{\sqrt[3]{{\frac{1}{{{n^6}}} - 1}}}^2}}}\\
= \frac{{\sqrt 1 + 1}}{{1 - 1.\sqrt[3]{{ - 1}} + {{\sqrt[3]{{ - 1}}}^2}}} = \frac{2}{3}
\end{array}\)
