# Câu 24 trang 152 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a.

\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{3{x^2} - x + 7} \over {2{x^3} - 1}} = \mathop {\lim }\limits_{x \to - \infty } {{{x^3}\left( {{3 \over x} - {1 \over {{x^2}}} + {7 \over {{x^3}}}} \right)} \over {{x^3}\left( {2 - {1 \over {{x^3}}}} \right)}} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{{3 \over x} - {1 \over {{x^2}}} + {7 \over {{x^3}}}} \over {2 - {1 \over {{x^3}}}}} = {0 \over 2} = 0 \cr}

b.

\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{2{x^4} + 7{x^3} - 15} \over {{x^4} + 1}} = \mathop {\lim }\limits_{x \to - \infty } {{{x^4}\left( {2 + {7 \over x} - {{15} \over {{x^4}}}} \right)} \over {{x^4}\left( {1 + {1 \over {{x^4}}}} \right)}} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{2 + {7 \over x} - {{15} \over {{x^4}}}} \over {1 + {1 \over {{x^4}}}}} = 2 \cr}

c.

\eqalign{ & \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^6} + 2} } \over {3{x^3} - 1}} = \mathop {\lim }\limits_{x \to - \infty } {{{x^3}\sqrt {1 + {2 \over {{x^6}}}} } \over {{x^3}\left( {3 - {1 \over {{x^3}}}} \right)}} \cr & = \mathop {\lim }\limits_{x \to - \infty } {{\sqrt {1 + {2 \over {{x^6}}}} } \over {3 - {1 \over {{x^3}}}}} = {1 \over 3} \cr}

d. Với mọi $$x < 0$$, ta có:

$${{\sqrt {{x^6} + 2} } \over {3{x^3} - 1}} = {{\left| x^3 \right|\sqrt {1 + {2 \over {{x^6}}}} } \over {3{x^3} - 1}} = {{ - {x^3}\sqrt {1 + {2 \over {{x^6}}}} } \over {3{x^3} - 1}} = {{ - \sqrt {1 + {2 \over {{x^6}}}} } \over {3 - {1 \over {{x^3}}}}}$$

Do đó :

$$\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^6} + 2} } \over {3{x^3} - 1}} = \mathop {\lim }\limits_{x \to - \infty } {{ - \sqrt {1 + {2 \over {{x^6}}}} } \over {3 - {1 \over {{x^3}}}}} = - {1 \over 3}$$