# Bài 6 trang 133 SGK Đại số và Giải tích 11

##### Hướng dẫn giải

$$\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right)$$

$$\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right)$$

$$\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right).g\left( x \right)$$

$$L > 0$$

$$+ \infty$$

$$+ \infty$$

$$- \infty$$

$$- \infty$$

$$L < 0$$

$$+ \infty$$

$$- \infty$$

$$- \infty$$

$$+ \infty$$

$$\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right)$$

$$\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right)$$

$$\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right).g\left( x \right)$$

$$L > 0$$

$$+ \infty$$

$$+ \infty$$

$$- \infty$$

$$- \infty$$

$$L < 0$$

$$+ \infty$$

$$- \infty$$

$$- \infty$$

$$+ \infty$$

Lời giải chi tiết

$$\begin{array}{l} a)\,\,\mathop {\lim }\limits_{x \to + \infty } \left( {{x^4} - {x^2} + x - 1} \right) \\= \mathop {\lim }\limits_{x \to + \infty } {x^4}\left( {1 - \frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} - \frac{1}{{{x^4}}}} \right)\\ \mathop {\lim }\limits_{x \to + \infty } {x^4} = + \infty \\ \mathop {\lim }\limits_{x \to + \infty } \left( {1 - \frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} - \frac{1}{{{x^4}}}} \right) = 1 > 0\\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( {{x^4} - {x^2} + x - 1} \right) = + \infty \\ b)\,\,\mathop {\lim }\limits_{x \to - \infty } \left( { - 2{x^3} + 3{x^2} - 5} \right) \\= \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 2 + \frac{1}{x} - \frac{5}{{{x^2}}}} \right)\\ \mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty \\ \mathop {\lim }\limits_{x \to - \infty } \left( { - 2 + \frac{1}{x} - \frac{5}{{{x^2}}}} \right) = - 2 < 0\\ \Leftrightarrow \mathop {\lim }\limits_{x \to - \infty } {x^3}\left( { - 2 + \frac{1}{x} - \frac{5}{{{x^2}}}} \right) = + \infty \\ c)\,\,\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} - 2x + 5} } \right) = \mathop {\lim }\limits_{x \to - \infty } \left| x \right|\sqrt {1 - \frac{2}{x} + \frac{5}{{{x^2}}}} \\ = \mathop {\lim }\limits_{x \to - \infty } \left[ { - x\sqrt {1 - \frac{2}{x} + \frac{5}{{{x^2}}}} } \right]\\ \mathop {\lim }\limits_{x \to - \infty } \left( { - x} \right) = + \infty \\ \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {1 - \frac{2}{x} + \frac{5}{{{x^2}}}} } \right) = 1 > 0\\ \Rightarrow \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} - 2x + 5} } \right) = + \infty \\ d)\,\,\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} + x}}{{5 - 2x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right)}}{{5 - 2x}}\\ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{1}{{{x^2}}}} + 1}}{{\frac{5}{x} - 2}} = \frac{{1 + 1}}{{ - 2}} = - 1 \end{array}$$