Bài 3 trang 79 SGK Toán 11 tập 1 - Chân trời sáng tạo

Đề bài

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{4x + 3}}{{2x}}\);      

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{3x + 1}}\);

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

Lời giải chi tiết

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{4x + 3}}{{2x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {4 + \frac{3}{x}} \right)}}{{2x}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{4 + \frac{3}{x}}}{2} = \frac{{\mathop {\lim }\limits_{x \to  + \infty } 4 + \mathop {\lim }\limits_{x \to  + \infty } \frac{3}{x}}}{{\mathop {\lim }\limits_{x \to  + \infty } 2}} = \frac{{4 + 0}}{2} = 2\)

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{3x + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{x\left( {3 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}.\mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{3 + \frac{1}{x}}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}.\frac{{\mathop {\lim }\limits_{x \to  - \infty } 2}}{{\mathop {\lim }\limits_{x \to  - \infty } 3 + \mathop {\lim }\limits_{x \to  - \infty } \frac{1}{x}}} = 0.\frac{2}{{3 + 0}} = 0\).

c) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {{x^2} + 1} }}{{x + 1}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\sqrt {1 + \frac{1}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}}\)

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