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