# Câu 42 trang 167 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a.  $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} + {1 \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} {{x + 1} \over {{x^2}}} = + \infty$$

vì  $$\mathop {\lim }\limits_{x \to 0} \left( {x + 1} \right) = 1 > 0,\mathop {\lim }\limits_{x \to 0} {x^2} = 0\,\text{ và }\,{x^2} > 0,\forall x \ne 0$$

b.

\eqalign{ & \mathop {\lim }\limits_{x \to - 2} {{{x^3} + 8} \over {x + 2}} = \mathop {\lim }\limits_{x \to - 2} {{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)} \over {x + 2}} \cr & = \mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 2x + 4} \right) = 12 \cr}

c.  $$\mathop {\lim }\limits_{x \to 9} {{3 - \sqrt x } \over {9 - x}} = \mathop {\lim }\limits_{x \to 9} {1 \over {3 + \sqrt x }} = {1 \over 6}$$

d.

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

e.

$$\mathop {\lim }\limits_{x \to + \infty } {{{x^4} - {x^3} + 11} \over {2x - 7}}$$

$$=\mathop {\lim }\limits_{x \to + \infty } {{{x^3} - {x^2} + {{11} \over x}} \over {2 - {7 \over x}}} = + \infty$$

f. Với $$x < 0$$, ta có :  $${{\sqrt {{x^4} + 4} } \over {x + 4}} = {{{x^2}\sqrt {1 + {4 \over {{x^4}}}} } \over {x + 4}} = {{x\sqrt {1 + {4 \over {{x^2}}}} } \over {1 + {4 \over x}}}$$

vì  $$\mathop {\lim }\limits_{x \to - \infty } x\sqrt {1 + {4 \over {{x^4}}}} = - \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to - \infty } \left( {1 + {4 \over x}} \right) = 1$$

nên $$\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^4} + 4} } \over {x + 4}} = - \infty$$