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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to – \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} – {x^2} + 3}}} \)
b. \(\mathop {\lim }\limits_{x \to – \infty } {{\left| x \right| + \sqrt {{x^2} + x} } \over {x + 10}}\)
c. \(\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} – 1} } \over {1 – 2x}}\)
d. \(\mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {2{x^2} + 1} + x} \right)\)
a. Với \(x < 0\), ta có :
\(\eqalign{
& x\sqrt {{{2{x^3} + x} \over {{x^5} – {x^2} + 3}}} = – \left| x \right|\sqrt {{{2{x^3} + x} \over {{x^5} – {x^2} + 3}}} \cr
& = – \sqrt {{{{x^2}\left( {2{x^3} + x} \right)} \over {{x^5} – {x^2} + 3}}} = – \sqrt {{{2 + {1 \over {{x^2}}}} \over {1 – {1 \over {{x^3}}} + {1 \over {{x^5}}}}}} \cr} \)
Do đó : \(\mathop {\lim }\limits_{x \to – \infty } x\sqrt {{{2{x^3} + x} \over {{x^5} – {x^2} + 3}}} = – \sqrt 2 \)
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b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \infty } {{\left| x \right|+\sqrt {{x^2} + x} } \over {x + 10}} = \mathop {\lim }\limits_{x \to – \infty } {{\left| x \right| + \left| x \right|\sqrt {1 + {1 \over x}} } \over {x + 10}} \cr
& = \mathop {\lim }\limits_{x \to – \infty } {{ – x – x\sqrt {1 + {1 \over x}} } \over {x + 10}} = \mathop {\lim }\limits_{x \to – \infty } {{ – 1 – \sqrt {1 + {1 \over x}} } \over {1 + {{10} \over x}}} \cr &= – 2 \cr} \)
c.
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2{x^4} + {x^2} – 1} } \over {1 – 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{{x^2}\sqrt {2 + {1 \over {{x^2}}} – {1 \over {{x^4}}}} } \over {x\left( {{1 \over x} – 2} \right)}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } x{{\sqrt {2 + {1 \over {{x^2}}} – {1 \over {{x^4}}}} } \over {{1 \over x} – 2}} = – \infty \cr
& \text{vì}\,\mathop {\lim }\limits_{x \to + \infty } x = + \infty \,\text{và}\,\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {2 + {1 \over {{x^2}}} – {1 \over {{x^4}}}} } \over {{1 \over x} – 2}} = – {{\sqrt 2 } \over 2} < 0 \cr} \)
d.
\(\eqalign{
& \mathop {\lim }\limits_{x \to – \infty } \left( {\sqrt {2{x^2} + 1} + x} \right) = \mathop {\lim }\limits_{x \to – \infty } {{2{x^2} + x – {x^2}} \over {\sqrt {2{x^2} + x} – x}} \cr
& = \mathop {\lim }\limits_{x \to – \infty } {{x\left( {x + 1} \right)} \over { – x\left( {\sqrt {2 + {1 \over x}} + 1} \right)}} \cr &= \mathop {\lim }\limits_{x \to – \infty } – {{x + 1} \over {\sqrt {2 + {1 \over x} + 1} }} = + \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to – \infty } \left( { – x – 1} \right) = + \infty \cr} \)