Bài 4 trang 72 Toán 11 tập 1 - Cánh Diều: Tính các giới hạn sau: $\mathop {\lim }\limits_{x \to + \infty } \frac{{9x + 1}}{{3x - 4}};$...

Tính các giới hạn sau:

a) $\mathop {\lim }\limits_{x \to + \infty } \frac{{9x + 1}}{{3x - 4}};$

b) $\mathop {\lim }\limits_{x \to - \infty } \frac{{7x - 11}}{{2x + 3}};$

c) $\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} }}{x};$

d) $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 1} }}{x};$

e) $\mathop {\lim }\limits_{x \to {6^ - }} \frac{1}{{x - 6}};$

g) $\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x - 7}}.$

- Sử dụng định lí về phép toán trên giới hạn hữu hạn của hàm số.

- Sử dụng giới hạn cơ bản sau: $\mathop {\lim }\limits_{x \to {a^ + }} \frac{1}{{x - a}} = + \infty ;\mathop {\lim }\limits_{x \to {a^ - }} \frac{1}{{x - a}} = - \infty$

a) $\mathop {\lim }\limits_{x \to + \infty } \frac{{9x + 1}}{{3x - 4}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {9 + \frac{1}{x}} \right)}}{{x\left( {3 - \frac{4}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{9 + \frac{1}{x}}}{{3 - \frac{4}{x}}} = \frac{{9 + 0}}{{3 - 0}} = 3$

b) $\mathop {\lim }\limits_{x \to - \infty } \frac{{7x - 11}}{{2x + 3}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\left( {7 - \frac{{11}}{x}} \right)}}{{x\left( {2 + \frac{3}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{7 - \frac{{11}}{x}}}{{2 + \frac{3}{x}}} = \frac{{7 - 0}}{{2 + 0}} = \frac{7}{2}$

c) $\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {1 + \frac{1}{{{x^2}}}} = \sqrt {1 + 0} = 1$

d) $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to - \infty } - \sqrt {1 + \frac{1}{{{x^2}}}} = - \sqrt {1 + 0} = - 1$

e) Ta có: $\left\{ \begin{array}{l}1 > 0\\x - 6 < 0,x \to {6^ - }\end{array} \right.$

Do đó, $\mathop {\lim }\limits_{x \to {6^ - }} \frac{1}{{x - 6}} = - \infty$

g) Ta có: $\left\{ \begin{array}{l}1 > 0\\x + 7 > 0,x \to {7^ + }\end{array} \right.$

Do đó, $\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x - 7}} = + \infty$