Bài 4.23 trang 165 SBT đại số và giải tích 11

LG a

$\mathop {\lim }\limits_{x \to  - 3} {{x + 3} \over {{x^2} + 2x - 3}}$

Lời giải chi tiết:

$\mathop {\lim }\limits_{x \to  - 3} {{x + 3} \over {{x^2} + 2x - 3}} = \mathop {\lim }\limits_{x \to  - 3} {{x + 3} \over {\left( {x - 1} \right)\left( {x + 3} \right)}} = \mathop {\lim }\limits_{x \to  - 3} {1 \over {x - 1}} = {{ - 1} \over 4}$

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LG b

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

Lời giải chi tiết:

$\mathop {\lim }\limits_{x \to  + \infty } {{x - 1} \over {{x^2} - 1}} = \mathop {\lim }\limits_{x \to  + \infty } {{{1 \over x} - {1 \over {{x^2}}}} \over {1 - {1 \over {{x^2}}}}} = 0$

LG c

$\mathop {\lim }\limits_{x \to 5} {{x - 5} \over {\sqrt x  - \sqrt 5 }}$

Lời giải chi tiết:

$\mathop {\lim }\limits_{x \to 5} {{x - 5} \over {\sqrt x  - \sqrt 5 }}$

$= \mathop {\lim }\limits_{x \to 5} {{\left( {\sqrt x  - \sqrt 5 } \right)\left( {\sqrt x  + \sqrt 5 } \right)} \over {\sqrt x  - \sqrt 5 }}$

$= \mathop {\lim }\limits_{x \to 5} \left( {\sqrt x  + \sqrt 5 } \right) = 2\sqrt 5$

LG d

$\mathop {\lim }\limits_{x \to  + \infty }  {{x - 5} \over {\sqrt x  + \sqrt 5 }}$

Lời giải chi tiết:

$\eqalign{ & \mathop {\lim }\limits_{x \to + \infty } {{x - 5} \over {\sqrt x + \sqrt 5 }} \cr  & = \mathop {\lim }\limits_{x \to + \infty } {{1 - {5 \over x}} \over {{1 \over {\sqrt x }} + {{\sqrt 5 } \over x}}} = + \infty \cr}$  

(Vì ${1 \over {\sqrt x }} + {{\sqrt 5 } \over x} > 0$ với mọi $x > 0$ )

LG e

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

Lời giải chi tiết:

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

LG f

$\mathop {\lim }\limits_{x \to  + \infty } {{1 - 2x + 3{x^3}} \over {{x^3} - 9}}$

Lời giải chi tiết:

$\mathop {\lim }\limits_{x \to  + \infty } {{1 - 2x + 3{x^3}} \over {{x^3} - 9}} = \mathop {\lim }\limits_{x \to  + \infty } {{{1 \over {{x^3}}} - {2 \over {{x^2}}} + 3} \over {1 - {9 \over {{x^3}}}}} = 3$

LG g

$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left( {{1 \over {{x^2} + 1}} - 1} \right)$

Lời giải chi tiết:

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

LG h

$\mathop {\lim }\limits_{x \to  - \infty } {{\left( {{x^2} - 1} \right){{\left( {1 - 2x} \right)}^5}} \over {{x^7} + x + 3}}$

Lời giải chi tiết:

$\eqalign{ & \mathop {\lim }\limits_{x \to - \infty } {{\left( {{x^2} - 1} \right){{\left( {1 - 2x} \right)}^5}} \over {{x^7} + x + 3}} \cr  & = \mathop {\lim }\limits_{x \to - \infty } {{{x^2}\left( {1 - {1 \over {{x^2}}}} \right).{x^5}{{\left( {{1 \over x} - 2} \right)}^5}} \over {{x^7} + x + 3}} \cr  & = \mathop {\lim }\limits_{x \to - \infty } {{\left( {1 - {1 \over {{x^2}}}} \right){{\left( {{1 \over x} - 2} \right)}^5}} \over {1 + {1 \over {{x^6}}} + {3 \over {{x^7}}}}} \cr  & = {\left( { - 2} \right)^5} = - 32 \cr}$

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