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