Đề bài

Tính các giới hạn sau: a) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{6x + 8}}{{5x - 2}}\);                      b) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{6x + 8}}{{5x - 2}}\);                  c) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}}\); d) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}}\);             e) \(\mathop {\lim }\limits_{x \to  - {2^ - }} \frac{{3{x^2} + 4}}{{2x + 4}}\);            g) \(\mathop {\lim }\limits_{x \to  - {2^ + }} \frac{{3{x^2} + 4}}{{2x + 4}}\).

Lời giải chi tiết

a) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{6x + 8}}{{5x - 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {6 + \frac{8}{x}} \right)}}{{x\left( {5 - \frac{2}{x}} \right)}} = \frac{6}{5}\)b) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{6x + 8}}{{5x - 2}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {6 + \frac{8}{x}} \right)}}{{x\left( {5 - \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{6 + \frac{8}{x}}}{{5 - \frac{2}{x}}} = \frac{6}{5}\).c) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {9 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {3 - \frac{2}{x}} \right)}} =  - \frac{3}{3} =  - 1\).d) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\sqrt {9 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {3 - \frac{2}{x}} \right)}} = \frac{3}{3} = 1\). e) \(\mathop {\lim }\limits_{x \to  - {2^ - }} \frac{{3{x^2} + 4}}{{2x + 4}} =  - \infty \)Do \(\mathop {\lim }\limits_{x \to  - {2^ - }} \left( {3{x^2} + 1} \right) = 3.{\left( { - 2} \right)^2} + 1 = 13 > 0\) và \(\mathop {\lim }\limits_{x \to  - {2^ - }} \frac{1}{{2x + 4}} =  - \infty \)g) \(\mathop {\lim }\limits_{x \to  - {2^ + }} \frac{{3{x^2} + 4}}{{2x + 4}} =  + \infty \).Do \(\mathop {\lim }\limits_{x \to  - {2^ + }} \left( {3{x^2} + 1} \right) = 3.{\left( { - 2} \right)^2} + 1 = 13 > 0\) và \(\mathop {\lim }\limits_{x \to  - {2^ + }} \frac{1}{{2x + 4}} =  + \infty \)