Bài 4 trang 72 SGK Toán 11 tập 1 - Cánh Diều

Đề bài

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

Lời giải chi tiết

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