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