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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to – \sqrt 3 } {{{x^3} + 3\sqrt 3 } \over {3 – {x^2}}}\)
b. \(\mathop {\lim }\limits_{x \to 4} {{\sqrt x – 2} \over {{x^2} – 4x}}\)
c. \(\mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x – 1} } \over {{x^2} – x}}\)
d. \(\mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^2} + x + 1} – 1} \over {3x}}\)
a. Ta có: \({{{x^3} + 3\sqrt 3 } \over {3 – {x^2}}} = {{\left( {x + \sqrt 3 } \right)\left( {{x^2} – x\sqrt 3 + 3} \right)} \over {\left( {x + \sqrt 3 } \right)\left( {\sqrt 3 – x} \right)}} = {{{x^2} – x\sqrt 3 + 3} \over {\sqrt 3 – x}}\)
với \(\,x \ne – \sqrt 3 \)
Do đó : \(\mathop {\lim }\limits_{x \to – \sqrt 3 } {{{x^3} + 3\sqrt 3 } \over {3 – {x^2}}} =\mathop {\lim }\limits_{x \to – \sqrt 3 } {{{x^2} – x\sqrt 3 + 3} \over {\sqrt 3 – x}}= {9 \over {2\sqrt 3 }} = {{3\sqrt 3 } \over 2}\)
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b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 4} {{\sqrt x – 2} \over {{x^2} – 4x}} = \mathop {\lim }\limits_{x \to 4} {{\sqrt x – 2} \over {x\left( {x – 4} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 4} {1 \over {x\left( {\sqrt x + 2} \right)}} = {1 \over {16}} \cr} \)
c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x – 1} } \over {{x^2} – x}} = \mathop {\lim }\limits_{x \to {1^ + }} {{\sqrt {x – 1} } \over {x\left( {x – 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} {1 \over {x\sqrt {x – 1} }} = + \infty \cr} \)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^2} + x + 1} – 1} \over {3x}} = \mathop {\lim }\limits_{x \to 0} {{{x^2} + x + 1 – 1} \over {3x(\sqrt {{x^2} + x + 1} + 1)}} \cr
& = {1 \over 3}\mathop {\lim }\limits_{x \to 0} {{x + 1} \over {\sqrt {{x^2} + x + 1} + 1}} = {1 \over 6} \cr} \)