Bài 4 trang 72 Toán 11 tập 1 – Cánh Diều: Tính các giới hạn sau: a) mathop lim limits_x -> + ∇ 9x + 1/3x – 4;

Đề bài/câu hỏi:

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) \(\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x – 7}}.\)

Hướng dẫn:

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

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

Lời giải:

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, \(\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x – 7}} = + \infty \)