\[\mathop {\lim }\limits_{x \to + \infty } \frac{{3x - 5\sin 2x + {{\cos }^2}x}}{{{x^2} + 2}}\] bằng:

\[\mathop {\lim }\limits_{x \to + \infty } \frac{{3x - 5\sin 2x + {{\cos }^2}x}}{{{x^2} + 2}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{3x}}{{{x^2} + 2}} - \mathop {\lim }\limits_{x \to + \infty } \frac{{5\sin 2x}}{{{x^2} + 2}} + \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\cos }^2}x}}{{{x^2} + 2}}\]

\[{A_1} = \mathop {\lim }\limits_{x \to + \infty } \frac{{3x}}{{{x^2} + 2}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{3}{x}}}{{1 + \frac{2}{{{x^2}}}}} = 0\].

\[\mathop {\lim }\limits_{x \to + \infty } \frac{{ - 5}}{{{x^2} + 2}} = 0 \le {A_2} = \mathop {\lim }\limits_{x \to + \infty } \frac{{5\sin 2x}}{{{x^2} + 2}} \le \mathop {\lim }\limits_{x \to + \infty } \frac{5}{{{x^2} + 2}} = 0 \Rightarrow {A_2} = 0\].

\[\mathop {\lim }\limits_{x \to + \infty } \frac{0}{{{x^2} + 2}} = 0 \le {A_3} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\cos }^2}x}}{{{x^2} + 2}} \le \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{x^2} + 2}} = 0 \Rightarrow {A_3} = 0\].

Vậy\[\mathop {\lim }\limits_{x \to + \infty } \frac{{3x - 5\sin 2x + {{\cos }^2}x}}{{{x^2} + 2}} = 0\]. Chọn B.