Tìm các giới hạn sau
a) \(\mathop {\lim }\limits_{x \to +\infty } {{2x - 3} \over {1 - 3x}}\) b) \(\mathop {\lim }\limits_{x \to -\infty } {{2{x^3} - 7{x^2} + 11} \over {3{x^6} + 2{x^5} - 5}}\)
c) \(\mathop {\lim }\limits_{x \to +\infty } x\sqrt {{{2x + 1} \over {3{x^3} + {x^2} + 2}}} \) d) \(\mathop {\lim }\limits_{x \to - \infty } {{2x + 3} \over {\sqrt {2{x^2} - 3} }}\)
Giải
a) \( - {2 \over 3}\) ; b) 0;
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c) \(x\root \of {{{2x + 1} \over {3{x^3} + {x^2} + 2}}} = \sqrt {{{{x^2}\left( {2x + 1} \right)} \over {3{x^3} + {x^2} + 2}}} \) với mọi \(x > 0\)
\(\mathop {\lim }\limits_{x \to + \infty }x \root \of {{{2x + 1} \over {3{x^3} + {x^2} + 2}}} = \sqrt {\mathop {\lim }\limits_{x \to + \infty } {{{x^2}\left( {2x + 1} \right)} \over {3{x^3} + {x^2} + 2}}} = \sqrt {{2 \over 3}} = {{\sqrt 6 } \over 3}\)
d) \(\mathop {\lim }\limits_{x \to - \infty } {{2x + 3} \over {\sqrt {2{x^2} - 3} }}= \mathop {\lim }\limits_{x \to - \infty } {{2x + 3} \over {|x|\sqrt {2 - {3 \over {{x^2}}}} }} \)
\(= \mathop {\lim }\limits_{x \to - \infty } {{2x + 3} \over { - x.\sqrt {2 - {3 \over {{x^2}}}} }} = \mathop {\lim }\limits_{x \to - \infty } {{2 + {3 \over x}} \over { - \sqrt {2 - {3 \over {{x^2}}}} }} = - \sqrt 2 \)