# Câu 31 trang 159 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a. Ta có:

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

b.

\eqalign{ & \mathop {\lim }\limits_{x \to 3} {{{x^4} - 27x} \over {2{x^2} - 3x - 9}} = \mathop {\lim }\limits_{x \to 3} {{x\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)} \over {\left( {x - 3} \right)\left( {2x + 3} \right)}} \cr & = \mathop {\lim }\limits_{x \to 3} {{x\left( {{x^2} + 3x + 9} \right)} \over {2x + 3}} = 9 \cr}

c.

\eqalign{ & \mathop {\lim }\limits_{x \to - 2} {{{x^4} - 16} \over {{x^2} + 6x + 8}} = \mathop {\lim }\limits_{x \to - 2} {{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)} \over {\left( {x + 2} \right)\left( {x + 4} \right)}} \cr & = \mathop {\lim }\limits_{x \to - 2} {{\left( {x - 2} \right)\left( {{x^2} + 4} \right)} \over {x + 4}} = - 16 \cr}

d.

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