Bài 4 trang 79 SGK Toán 11 tập 1 – Cánh Diều

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 \(\mathop {\lim }\limits_{x \to  – {2^ + }} \frac{1}{{2x + 4}} =  + \infty \)