Bài tập 25 trang 152 SGK Toán 11 NC

a)

\(\mathop {\lim }\limits_{x \to  - \infty } \sqrt[3]{{\frac{{{x^2} + 2x}}{{8{x^2} - x + 3}}}} = \mathop {\lim }\limits_{x \to  - \infty } \sqrt[3]{{\frac{{1 + \frac{2}{x}}}{{8 - \frac{1}{x} + \frac{3}{{{x^2}}}}}}} = \frac{1}{2}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \frac{{x\sqrt x }}{{{x^2} - x + 2}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\sqrt x }}{{{x^2}\left( {1 - \frac{1}{x} + \frac{2}{{{x^2}}}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt x \left( {1 - \frac{1}{x} + \frac{2}{{{x^2}}}} \right)}} = 0
\end{array}\)

(vì \(\mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt x }} = 0,\mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{1 - \frac{1}{x} + \frac{2}{{{x^2}}}}} = 1\))

-- Mod Toán 11 HỌC247