Tìm các giới hạn sau
a) lim
b) \mathop {\lim }\limits_{x \to - \infty } {{{x^2} - 5x + 2} \over {2\left| x \right| + 1}}
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Giải
\eqalign{ & a)\,\,\mathop {\lim }\limits_{x \to - \infty } {{2 - {1 \over {{x^3}}} - {1 \over {{x^4}}}} \over {{1 \over {{x^2}}} + {1 \over {{x^3}}} + {2 \over {{x^4}}}}} = + \infty \cr & b)\,\mathop {\lim }\limits_{x \to - \infty } {{{x^2} - 5x + 2} \over { - 2x + 1}} = \mathop {\lim }\limits_{x \to - \infty } {{1 - {5 \over x} + {2 \over {{x^2}}}} \over { - {2 \over x} + {1 \over {{x^2}}}}}\cr&= + \infty \cr}