Bài 4 trang 72 SGK Toán 11 Cánh diều Tập 1

Tính các giới hạn sau:

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

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

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

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

e) $\lim_{x \to 7^{+}} \frac{1}{x - 7}$ .

QUẢNG CÁO

Hướng dẫn giải chi tiết Bài 4

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

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

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

d) $\underset{x\to -\infty }{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+1}}{x}=\underset{x\to +\infty }{\mathop{\lim }}\,\frac{\left| x \right|\sqrt{1+\frac{1}{{{x}^{2}}}}}{x}=-1$

e) $\underset{x\to {{7}^{+}}}{\mathop{\lim }}\,\frac{1}{x-7}=+\infty $

-- Mod Toán 11 HỌC247

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