Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} - 5}}{{{x^2} + 1}}\)
b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4} - x} }}{{1 - 2x}}\)
a)
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} - 5}}{{{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } x.\frac{{{x^2}\left( {1 - \frac{5}{{{x^3}}}} \right)}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}}\\
= \mathop {\lim }\limits_{x \to + \infty } x.\frac{{1 - \frac{5}{{{x^3}}}}}{{1 + \frac{1}{{{x^2}}}}} = + \infty
\end{array}\)
(vì \(\mathop {\lim }\limits_{x \to + \infty } x = + \infty ,\mathop {\lim }\limits_{x \to + \infty } \frac{{1 - \frac{5}{{{x^3}}}}}{{1 + \frac{1}{{{x^2}}}}} = 1 > 0\))
b) Với mọi x < 0, ta có:
\(\frac{{\sqrt {{x^4} - x} }}{{1 - 2x}} = \frac{{{x^2}\sqrt {1 - \frac{1}{{{x^3}}}} }}{{1 - 2x}} = \frac{{\sqrt {1 - \frac{1}{{{x^3}}}} }}{{\frac{1}{{{x^2}}} - \frac{2}{x}}}\)
Vì \(\mathop {\lim }\limits_{x \to - \infty } \sqrt {1 - \frac{1}{{{x^3}}}} = 1,\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{1}{{{x^2}}} - \frac{2}{x}} \right) = 0\) và \(\frac{1}{{{x^2}}} - \frac{2}{x} > 0\) với x < 0)
Do đó \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^4} - x} }}{{1 - 2x}} = + \infty \)
-- Mod Toán 11