# Câu 32 trang 159 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a.  $$\mathop {\lim }\limits_{x \to + \infty } \root 3 \of {{{2{x^5} + {x^3} - 1} \over {\left( {2{x^2} - 1} \right)\left( {{x^3} + x} \right)}}} = \mathop {\lim }\limits_{x \to + \infty } \root 3 \of {{{2 + {1 \over {{x^2}}} - {1 \over {{x^5}}}} \over {\left( {2 - {1 \over {{x^2}}}} \right)\left( {1 + {1 \over {{x^2}}}} \right)}}} = 1$$

b.

\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{2\left| x \right| + 3} \over {\sqrt {{x^2} + x + 5} }} = \mathop {\lim }\limits_{x \to - \infty } {{2\left| x \right| + 3} \over {\left| x \right|\sqrt {1 + {1 \over x} + {5 \over {{x^2}}}} }} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{ - 2x + 3} \over { - x\sqrt {1 + {1 \over x} + {5 \over {{x^2}}}} }} =\mathop {\lim }\limits_{x \to - \infty } {{2 - {3 \over x}} \over {\sqrt {1 + {1 \over x} + {5 \over {{x^2}}}} }}= 2 \cr}

c.  $${x^2} + x \ge 0 \Leftrightarrow x \le - 1\,\text{ hoặc }\,x \ge 0$$

Với mọi $$x ≤ -1$$,  $$x \ne - {3 \over 2}$$

$${{\sqrt {{x^2} + x} + 2x} \over {2x + 3}} = {{\left| x \right|\sqrt {1 + {1 \over x}} + 2x} \over {2x + 3}} = {{ - x\sqrt {1+ {1 \over x}} + 2x} \over {2x + 3}} = {{ - \sqrt {1 + {1 \over x}} + 2} \over {2 + {3 \over x}}}$$

Do đó  $$\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^2} + x} + 2x} \over {2x + 3}} =\mathop {\lim }\limits_{x \to - \infty }{{ - \sqrt {1 + {1 \over x}} + 2} \over {2 + {3 \over x}}}= {1 \over 2}$$

d.

\eqalign{ & \mathop {\lim }\limits_{x \to + \infty } \left( {x + 1} \right)\sqrt {{x \over {2{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{x{{\left( {x + 1} \right)}^2}} \over {2{x^4} + {x^2} + 1}}} \cr & = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{1 \over x} + {2 \over {{x^2}}} + {1 \over {{x^3}}}} \over {2 + {1 \over {{x^2}}} + {1 \over {{x^4}}}}}} = 0 \cr}