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