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Bài 25. Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to – \infty } \root 3 \of {{{{x^2} + 2x} \over {8{x^2} – x + 3}}} \)
b. \(\mathop {\lim }\limits_{x \to + \infty } {{x\sqrt x } \over {{x^2} – x + 2}}\)
a. Ta có:
\(\mathop {\lim }\limits_{x \to – \infty } \root 3 \of {{{{x^2} + 2x} \over {8{x^2} – x + 3}}} = \mathop {\lim }\limits_{x \to – \infty } \root 3 \of {{{1 + {2 \over x}} \over {8 – {1 \over x} + {3 \over {{x^2}}}}}} = {1 \over 2}\)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{x\sqrt x } \over {{x^2} – x + 2}} = \mathop {\lim }\limits_{x \to + \infty } {{x\sqrt x } \over {{x^2}\left( {1 – {1 \over x} + {2 \over {{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } {1 \over {\sqrt x \left( {1 – {1 \over x} + {2 \over {{x^2}}}} \right)}} = 0 \cr
& \text{vì}\;\mathop {\lim }\limits_{x \to + \infty } {1 \over {\sqrt x }} = 0\;\text{và}\;\mathop {\lim }\limits_{x \to + \infty } {1 \over {1 – {1 \over x} + {2 \over {{x^2}}}}} = 1 \cr} \)