Sử dụng các định lí về giới hạn hàm số. Phân tích, đưa ra lời giải Giải bài 21 trang 76 sách bài tập toán 11 – Cánh diều – Bài 2. Giới hạn của hàm số. Tính các giới hạn sau:…
Đề bài/câu hỏi:
Tính các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to – \infty } \frac{{ – 5x + 2}}{{3x + 1}}\)
b) \(\mathop {\lim }\limits_{x \to – \infty } \frac{{ – 2x + 3}}{{3{x^2} + 2x + 5}}\)
c) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}}\)
d) \(\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}}\)
e) \(\mathop {\lim }\limits_{x \to 1} \frac{{2{x^2} – 8x + 6}}{{{x^2} – 1}}\)
g) \(\mathop {\lim }\limits_{x \to – 3} \frac{{ – {x^2} + 2x + 15}}{{{x^2} + 4x + 3}}\)
Hướng dẫn:
Sử dụng các định lí về giới hạn hàm số.
Lời giải:
a) Ta có:\(\mathop {\lim }\limits_{x \to – \infty } \frac{{ – 5x + 2}}{{3x + 1}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{x\left( { – 5 + \frac{2}{x}} \right)}}{{x\left( {3 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – 5 + \frac{2}{x}}}{{3 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to – \infty } \left( { – 5} \right) + \mathop {\lim }\limits_{x \to – \infty } \frac{2}{x}}}{{\mathop {\lim }\limits_{x \to – \infty } 3 + \mathop {\lim }\limits_{x \to – \infty } \frac{1}{x}}}\)
\( = \frac{{ – 5 + 0}}{{3 + 0}} = \frac{{ – 5}}{3}\)
b) Ta có: \(\mathop {\lim }\limits_{x \to – \infty } \frac{{ – 2x + 3}}{{3{x^2} + 2x + 5}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{{x^2}\left( {\frac{{ – 2}}{x} + \frac{3}{{{x^2}}}} \right)}}{{{x^2}\left( {3 + \frac{2}{x} + \frac{5}{{{x^2}}}} \right)}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\frac{{ – 2}}{x} + \frac{3}{{{x^2}}}}}{{3 + \frac{2}{x} + \frac{5}{{{x^2}}}}}\)
\( = \frac{{\mathop {\lim }\limits_{x \to – \infty } \frac{{ – 2}}{x} + \mathop {\lim }\limits_{x \to – \infty } \frac{3}{{{x^2}}}}}{{\mathop {\lim }\limits_{x \to – \infty } 3 + \mathop {\lim }\limits_{x \to – \infty } \frac{2}{x} + \mathop {\lim }\limits_{x \to – \infty } \frac{5}{{{x^2}}}}} = \frac{{0 + 0}}{{3 + 0 + 0}} = 0\).
c) Ta có:
\(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2}\left( {9 + \frac{3}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {9 + \frac{3}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9 + \frac{3}{{{x^2}}}} }}{{1 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } \sqrt {9 + \frac{3}{{{x^2}}}} }}{{\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right)}}\)
Do \(\mathop {\lim }\limits_{x \to + \infty } \left( {9 + \frac{3}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to + \infty } 9 + \mathop {\lim }\limits_{x \to + \infty } \frac{3}{{{x^2}}} = 9 + 0 = 9\), nên \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {9 + \frac{3}{{{x^2}}}} = \sqrt 9 = 3\).
Mặt khác, \(\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x} = 1 + 0 = 1\).
Suy ra \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } \sqrt {9 + \frac{3}{{{x^2}}}} }}{{\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right)}} = \frac{3}{1} = 3\).
d) Ta có: \(\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {{x^2}\left( {9 + \frac{3}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( { – x} \right)\sqrt {9 + \frac{3}{{{x^2}}}} }}{{x\left( {1 + \frac{1}{x}} \right)}}\)
\( = \mathop {\lim }\limits_{x \to – \infty } \left( { – \frac{{\sqrt {9 + \frac{3}{{{x^2}}}} }}{{1 + \frac{1}{x}}}} \right) = – \frac{{\mathop {\lim }\limits_{x \to – \infty } \sqrt {9 + \frac{3}{{{x^2}}}} }}{{\mathop {\lim }\limits_{x \to – \infty } \left( {1 + \frac{1}{x}} \right)}}\)
Do \(\mathop {\lim }\limits_{x \to – \infty } \left( {9 + \frac{3}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to – \infty } 9 + \mathop {\lim }\limits_{x \to – \infty } \frac{3}{{{x^2}}} = 9 + 0 = 9\), nên \(\mathop {\lim }\limits_{x \to – \infty } \sqrt {9 + \frac{3}{{{x^2}}}} = \sqrt 9 = 3\).
Mặt khác, \(\mathop {\lim }\limits_{x \to – \infty } \left( {1 + \frac{1}{x}} \right) = \mathop {\lim }\limits_{x \to – \infty } 1 + \mathop {\lim }\limits_{x \to – \infty } \frac{1}{x} = 1 + 0 = 1\).
Suy ra \(\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {9{x^2} + 3} }}{{x + 1}} = – \frac{{\mathop {\lim }\limits_{x \to – \infty } \sqrt {9 + \frac{3}{{{x^2}}}} }}{{\mathop {\lim }\limits_{x \to – \infty } \left( {1 + \frac{1}{x}} \right)}} = – \frac{3}{1} = – 3\).
e) Ta có: \(\mathop {\lim }\limits_{x \to 1} \frac{{2{x^2} – 8x + 6}}{{{x^2} – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x – 1} \right)\left( {2x – 6} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{2x – 6}}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} 2x – \mathop {\lim }\limits_{x \to 1} 6}}{{\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1}}\)
\( = \frac{{2.1 – 6}}{{1 + 1}} = – 2\).
f) Ta có: \(\mathop {\lim }\limits_{x \to – 3} \frac{{ – {x^2} + 2x + 15}}{{{x^2} + 4x + 3}} = \mathop {\lim }\limits_{x \to – 3} \frac{{\left( {x + 3} \right)\left( {5 – x} \right)}}{{\left( {x + 3} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to – 3} \frac{{5 – x}}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to – 3} 5 – \mathop {\lim }\limits_{x \to – 3} x}}{{\mathop {\lim }\limits_{x \to – 3} x + \mathop {\lim }\limits_{x \to – 3} 1}}\)
\( = \frac{{5 – \left( { – 3} \right)}}{{\left( { – 3} \right) + 1}} = – 4\).