Đáp án đúng: C
Giải chi tiết:Dựa vào đáp án, ta thấy rằng:
\(y={{x}^{3}}-x-1\,\,\xrightarrow{{}}\,\,\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,y=\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,\left( {{x}^{3}}-x-1 \right)=\infty \,\,\Rightarrow \) ĐTHS không có TCN. \(y=\frac{{{x}^{3}}+1}{{{x}^{2}}+1}\,\,\xrightarrow{{}}\,\,\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,y=\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,\frac{{{x}^{3}}+1}{{{x}^{2}}+1}=\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,\frac{1+\frac{1}{{{x}^{3}}}}{\frac{1}{x}+\frac{1}{{{x}^{3}}}}=\infty \,\,\Rightarrow \) ĐTHS không có TCN. \(y=\frac{3{{x}^{2}}+2x-1}{4{{x}^{2}}+5}\,\,\xrightarrow{{}}\,\,\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,y=\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}+2x-1}{4{{x}^{2}}+5}=\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,\frac{3+\frac{2}{x}-\frac{1}{{{x}^{2}}}}{4+\frac{5}{{{x}^{2}}}}=\frac{3}{4}\)\(\Rightarrow y=\frac{3}{4}\) là TCN. \(y=\sqrt{2{{x}^{2}}+3}\,\,\xrightarrow{{}}\,\,\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,y=\underset{x\,\to \,\,\infty }{\mathop{\lim }}\,\sqrt{2{{x}^{2}}+3}=\infty \,\,\Rightarrow \) ĐTHS không có TCN.
Chọn C
