Question

[Quoc Tran Anh Le]

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

Answer 1

[Hà Quang Minh]

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

Answer 2

[Hà Quang Minh]

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