\(\eqalign{& \mathop {\lim }\limits_{x \to - \sqrt 2 } = {{{x^3} + 2\sqrt 2 } \over {{x^2} - 2}}\cr & = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^3} + {{\left( {\sqrt 2 } \right)}^3}} \over {{x^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr& = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{\left( {x + \sqrt 2 } \right)\left( {{x^2} - x\sqrt 2 + 2} \right)} \over {\left( {x + \sqrt 2 } \right)\left( {x - \sqrt 2 } \right)}} \cr& = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^2} - x\sqrt 2 + 2} \over {x - \sqrt 2 }} \cr &= \frac{{{{\left( { - \sqrt 2 } \right)}^2} - \left( { - \sqrt 2 } \right).\sqrt 2 + 2}}{{ - \sqrt 2 - \sqrt 2 }}\cr &= {{ - 3\sqrt 2 } \over 2} \cr} \)
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LG a
\(\mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^3} + 2\sqrt 2 } \over {{x^2} - 2}}\) Phương pháp giải: Phân tích tử và mẫu của phân thức thành nhân tử, khử dạng vô định và tính giới hạn. Lời giải chi tiết: Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to - \sqrt 2 } = {{{x^3} + 2\sqrt 2 } \over {{x^2} - 2}}\cr & = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^3} + {{\left( {\sqrt 2 } \right)}^3}} \over {{x^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{\left( {x + \sqrt 2 } \right)\left( {{x^2} - x\sqrt 2 + 2} \right)} \over {\left( {x + \sqrt 2 } \right)\left( {x - \sqrt 2 } \right)}} \cr
& = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^2} - x\sqrt 2 + 2} \over {x - \sqrt 2 }} \cr &= \frac{{{{\left( { - \sqrt 2 } \right)}^2} - \left( { - \sqrt 2 } \right).\sqrt 2 + 2}}{{ - \sqrt 2 - \sqrt 2 }}\cr &= {{ - 3\sqrt 2 } \over 2} \cr} \)
LG b
\(\mathop {\lim }\limits_{x \to 3} {{{x^4} - 27x} \over {2{x^2} - 3x - 9}}\) Lời giải chi tiết: \(\eqalign{
& \mathop {\lim }\limits_{x \to 3} {{{x^4} - 27x} \over {2{x^2} - 3x - 9}} \cr &= \mathop {\lim }\limits_{x \to 3} \frac{{x\left( {{x^3} - 27} \right)}}{{\left( {x - 3} \right)\left( {2x + 3} \right)}}\cr &= \mathop {\lim }\limits_{x \to 3} {{x\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)} \over {\left( {x - 3} \right)\left( {2x + 3} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 3} {{x\left( {{x^2} + 3x + 9} \right)} \over {2x + 3}} \cr &= \frac{{3\left( {{3^2} + 3.3 + 9} \right)}}{{2.3 + 3}}= 9 \cr} \)
LG c
\(\mathop {\lim }\limits_{x \to - 2} {{{x^4} - 16} \over {{x^2} + 6x + 8}}\) Lời giải chi tiết: \(\eqalign{
& \mathop {\lim }\limits_{x \to - 2} {{{x^4} - 16} \over {{x^2} + 6x + 8}} \cr &= \mathop {\lim }\limits_{x \to - 2} {{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)} \over {\left( {x + 2} \right)\left( {x + 4} \right)}} \cr
&= \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}{{\left( {x + 2} \right)\left( {x + 4} \right)}}\cr &= \mathop {\lim }\limits_{x \to - 2} {{\left( {x - 2} \right)\left( {{x^2} + 4} \right)} \over {x + 4}} \cr &= \frac{{\left( { - 2 - 2} \right)\left( {{{\left( { - 2} \right)}^2} + 4} \right)}}{{ - 2 + 4}}= - 16 \cr} \)
LG d
\(\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt {1 - x} + x - 1} \over {\sqrt {{x^2} - {x^3}} }}\) Lời giải chi tiết: \(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt {1 - x} + x - 1} \over {\sqrt {{x^2} - {x^3}} }} \cr &= \mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt {1 - x} - \left( {1 - x} \right)} \over {{\sqrt {{x^2}\left( {1 - x} \right)} }}} \cr
&= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x} \left( {1 - \sqrt {1 - x} } \right)}}{{\left| x \right|\sqrt {1 - x} }}\cr &= \mathop {\lim }\limits_{x \to {1^ - }} {{1 - \sqrt {1 - x} } \over {\left| x \right|}} \cr &= \frac{{1 - \sqrt {1 - 1} }}{{\left| 1 \right|}}= 1 \cr} \)
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