Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^3} + 2\sqrt 2 } \over {{x^2} - 2}}\)
b. \(\mathop {\lim }\limits_{x \to 3} {{{x^4} - 27x} \over {2{x^2} - 3x - 9}}\)
c. \(\mathop {\lim }\limits_{x \to - 2} {{{x^4} - 16} \over {{x^2} + 6x + 8}}\)
d. \(\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt {1 - x} + x - 1} \over {\sqrt {{x^2} - {x^3}} }}\)
a. Ta có:
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\(\eqalign{
& \mathop {\lim }\limits_{x \to - \sqrt 2 } = {{{x^3} + 2\sqrt 2 } \over {{x^2} - 2}} = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^3} + {{\left( {\sqrt 2 } \right)}^3}} \over {{x^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr
& = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{\left( {x + \sqrt 2 } \right)\left( {{x^2} - x\sqrt 2 + 2} \right)} \over {\left( {x + \sqrt 2 } \right)\left( {x - \sqrt 2 } \right)}} \cr
& = \mathop {\lim }\limits_{x \to - \sqrt 2 } {{{x^2} - x\sqrt 2 + 2} \over {x - \sqrt 2 }} = {{ - 3\sqrt 2 } \over 2} \cr} \)
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.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt {1 - x} + x - 1} \over {\sqrt {{x^2} - {x^3}} }} = \mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt {1 - x} - \left( {1 - x} \right)} \over {\left| x \right|\sqrt {1 - x} }} \cr
& = \mathop {\lim }\limits_{x \to {1^ - }} {{1 - \sqrt {1 - x} } \over {\left| x \right|}} = 1 \cr} \)