Advertisements (Quảng cáo)
Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to – \infty } \left( {3{x^3} – 5{x^2} + 7} \right)\)
b. \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {2{x^4} – 3x + 12} \)
a. Ta có:
\(\mathop {\lim }\limits_{x \to – \infty } \left( {3{x^3} – 5{x^2} + 7} \right) = \mathop {\lim }\limits_{x \to – \infty } {x^3}\left( {3 – {5 \over x} + {7 \over {{x^3}}}} \right) = – \infty \)
Advertisements (Quảng cáo)
Vì \(\mathop {\lim }\limits_{x \to – \infty } {x^3} = – \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to – \infty } \left( {3 – {5 \over x} + {7 \over {{x^3}}}} \right) = 3 > 0\)
b.
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } \sqrt {2{x^4} – 3x + 12} = \mathop {\lim }\limits_{x \to + \infty } {x^2}\sqrt {2 – {3 \over {{x^3}}} + {{12} \over {{x^4}}}} = + \infty \cr
& \text{vì }\,\mathop {\lim }\limits_{x \to + \infty } {x^2} = + \infty \,\text{ và }\,\mathop {\lim }\limits_{x \to + \infty } \sqrt {2 – {3 \over {{x^3}}} + {{12} \over {{x^4}}}} = \sqrt 2 > 0 \cr} \)