Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to + \infty } {{\left( {2x - 5} \right){{\left( {1 - x} \right)}^2}} \over {3{x^3} - x + 1}}\) b) \(\mathop {\lim }\limits_{x \to - \infty } {{\left( {2x - 1} \right)\sqrt {{x^2} - 3} } \over {x - 5{x^2}}}\)
c) \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {{{{x^4} +{x^2} + 2} \over {\left( {{x^3} + 1} \right)\left( {3x - 1} \right)}}} \) d) \(\mathop {\lim }\limits_{x \to - \infty } {{2x - 3} \over {\sqrt {{x^2} + 1} - x}}.\)
a) \({2 \over 3};\) b) \({2 \over 5};\) c) \({{\sqrt 3 } \over 3};\)
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d) Với mọi \(x < 0,\) ta có
\({{2x - 3} \over {\sqrt {{x^2} + 1} - x}} = {{2x - 3} \over {\left| x \right|\sqrt {1 + {1 \over {{x^2}}}} - x}} = {{2x - 3} \over { - x\sqrt {1 + {1 \over {{x^2}}}} - x}} = {{2 - {3 \over x}} \over { - \sqrt {1 + {1 \over {{x^2}}}} - 1}}\)
Do đó
\(\mathop {\lim }\limits_{x \to - \infty } {{2x - 3} \over {\sqrt {{x^2} + 1} - x}} = - 1.\)