Bài tập 32 trang 159 SGK Toán 11 NC

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \sqrt[3]{{\frac{{2{x^5} + {x^3} - 1}}{{\left( {2{x^2} - 1} \right)\left( {{x^3} + x} \right)}}}}\\
 = \sqrt[3]{{\frac{{2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^5}}}}}{{\left( {2 - \frac{1}{{{x^2}}}} \right)\left( {1 + \frac{1}{{{x^2}}}} \right)}}}} = 1
\end{array}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{2\left| x \right| + 3}}{{\sqrt {{x^2} + x + 5} }} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{2\left| x \right| + 3}}{{\left| x \right|\sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - 2x + 3}}{{ - x\sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{2 - \frac{3}{x}}}{{\sqrt {1 + \frac{1}{x} + \frac{5}{{{x^2}}}} }} = 2
\end{array}\)

c) \({x^2} + x \ge 0 \Leftrightarrow x \le  - 1 \vee x \ge 0\)

\(\begin{array}{l}
\frac{{\sqrt {{x^2} + x}  + 2x}}{{2x + 3}} = \frac{{\left| x \right|\sqrt {1 + \frac{1}{x}}  + 2x}}{{2x + 3}}\\
 = \frac{{ - \sqrt {1 + \frac{1}{x}}  + 2x}}{{2 + \frac{3}{x}}}
\end{array}\)

Với mọi \(x \le  - 1,x \ne  - \frac{3}{2}\)

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

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {x + 1} \right)\sqrt {\frac{x}{{2{x^4} + {x^2} + 1}}} \\
 = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{x{{\left( {x + 1} \right)}^2}}}{{2{x^4} + {x^2} + 1}}} 
\end{array}\\
{ = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {\frac{{\frac{1}{x} + \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}}}{{2 + \frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}}  = 0}
\end{array}\)

-- Mod Toán 11 HỌC247