Đáp án:
\[\mathop {\lim }\limits_{x \to + \infty } \frac{{3{x^2} - 2x\sqrt x + 4}}{{{{\left( {2x + 1} \right)}^2}}} = \frac{3}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \frac{{3{x^2} - 2x\sqrt x + 4}}{{{{\left( {2x + 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{3{x^2} - 2x\sqrt x + 4}}{{4{x^2} + 4x + 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{3 - \frac{2}{{\sqrt x }} + \frac{4}{{{x^2}}}}}{{4 + \frac{4}{x} + \frac{1}{{{x^2}}}}}\\
= \frac{{3 - 0 + 0}}{{4 + 0 + 0}}\\
= \frac{3}{4}
\end{array}\)