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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to + \infty } {{2{x^2} + x – 10} \over {9 – 3{x^3}}}\)
b. \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {2{x^2} – 7x + 12} } \over {3\left| x \right| – 17}}\)
a. \(\mathop {\lim }\limits_{x \to + \infty } {{2{x^2} + x – 10} \over {9 – 3{x^3}}} = \mathop {\lim }\limits_{x \to + \infty } {{{2 \over x} + {1 \over {{x^2}}} – {{10} \over {{x^3}}}} \over {{9 \over {{x^3}}} – 3}} = 0\)
Advertisements (Quảng cáo)
b. Với mọi \(x ≠ 0\), ta có :
\({{\sqrt {2{x^2} – 7x + 12} } \over {3\left| x \right| – 17}} = {{\left| x \right|\sqrt {2 – {7 \over x} + {{12} \over {{x^2}}}} } \over {\left| x \right|\left( {3 – {{17} \over {\left| x \right|}}} \right)}} = {{\sqrt {2 – {7 \over x} + {{12} \over {{x^2}}}} } \over {3 – {{17} \over {\left| x \right|}}}}\)
Do đó \(\mathop {\lim }\limits_{x \to – \infty } {{\sqrt {2{x^2} – 7x + 12} } \over {3\left| x \right| – 17}} = {{\sqrt 2 } \over 3}\)