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