Đáp án đúng: D Ta có $\lim \left( 3{{n}^{4}}+4{{n}^{2}}-n+1 \right)=\lim {{n}^{4}}\left( 3+\frac{4}{{{n}^{2}}}-\frac{1}{{{n}^{3}}}+\frac{1}{{{n}^{4}}} \right)=+\infty $ vì$\left\{ \begin{array}{l}\lim {{n}^{4}}=+\infty \\\lim \left( 3+\frac{4}{{{n}^{2}}}-\frac{1}{{{n}^{3}}}+\frac{1}{{{n}^{4}}} \right)=3>0\end{array} \right..$ Chọn D. Giải nhanh: $3{{n}^{4}}+4{{n}^{2}}-n+1\sim 3{{n}^{4}}\xrightarrow[{}]{}+\infty .$