Sử dụng các tính chất về giới hạn hàm số. Gợi ý giải Giải bài 44 trang 83 sách bài tập toán 11 – Cánh diều – Bài tập cuối Chương 3. Tính các giới hạn sau:…
Đề bài/câu hỏi:
Tính các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to – \infty } \frac{{2 + \frac{4}{{3x}}}}{{{x^2} – 1}}\)
b) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x – 2}}\)
c) \(\mathop {\lim }\limits_{x \to – {3^ + }} \frac{{ – 5 + x}}{{x + 3}}\)
d) \(\mathop {\lim }\limits_{x \to – \infty } \frac{{14x + 2}}{{ – 7x + 1}}\)
e) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ – 2{x^2}}}{{3x + 5}}\)
g) \(\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {4{x^2} + 1} }}{{x + 2}}\)
h) \(\mathop {\lim }\limits_{x \to 1} \frac{{x – 1}}{{{x^2} – 1}}\)
i) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 5x + 6}}{{x – 2}}\)
k) \(\mathop {\lim }\limits_{x \to 3} \frac{{ – {x^2} + 4x – 3}}{{{x^2} + 3x – 18}}\)
Hướng dẫn:
Sử dụng các tính chất về giới hạn hàm số.
Lời giải:
a) Ta có \(\mathop {\lim }\limits_{x \to – \infty } \left( {2 + \frac{4}{{3x}}} \right) = \mathop {\lim }\limits_{x \to – \infty } 2 + \mathop {\lim }\limits_{x \to – \infty } \frac{4}{{3x}} = 2 + 0 = 2\).
Mặt khác, \(\mathop {\lim }\limits_{x \to – \infty } \left( {{x^2} – 1} \right) = \mathop {\lim }\limits_{x \to – \infty } \left[ {{x^2}\left( {1 – \frac{1}{{{x^2}}}} \right)} \right] = \mathop {\lim }\limits_{x \to – \infty } {x^2}.\mathop {\lim }\limits_{x \to – \infty } \left( {1 – \frac{1}{{{x^2}}}} \right) = + \infty \)
Suy ra \(\mathop {\lim }\limits_{x \to – \infty } \frac{{2 + \frac{4}{{3x}}}}{{{x^2} – 1}} = 0\).
b) Ta có \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x – 2}} = + \infty \).
c) Ta có \(\mathop {\lim }\limits_{x \to – {3^ + }} \left( { – 5 + x} \right) = \left( { – 5} \right) + \left( { – 3} \right) = – 2 < 0\).
Suy ra \(\mathop {\lim }\limits_{x \to – {3^ + }} \frac{{ – 5 + x}}{{x + 3}} = – \infty \).
d) Ta có:\(\mathop {\lim }\limits_{x \to – \infty } \frac{{14x + 2}}{{ – 7x + 1}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{x\left( {14 + \frac{2}{x}} \right)}}{{x\left( { – 7 + \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{14 + \frac{2}{x}}}{{ – 7 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to – \infty } 14 + \mathop {\lim }\limits_{x \to – \infty } \frac{2}{x}}}{{\mathop {\lim }\limits_{x \to – \infty } \left( { – 7} \right) + \mathop {\lim }\limits_{x \to – \infty } \frac{1}{x}}}\)
\( = \frac{{14 + 0}}{{ – 7 + 0}} = – 2\).
e) Ta có \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ – 2{x^2}}}{{3x + 5}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ – 2{x^2}}}{{x\left( {3 + \frac{5}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{ – 2x}}{{3 + \frac{5}{x}}}\).
Ta thấy \(\mathop {\lim }\limits_{x \to + \infty } \left( { – 2x} \right) = – \infty \) và \(\mathop {\lim }\limits_{x \to + \infty } \left( {3 + \frac{5}{x}} \right) = \mathop {\lim }\limits_{x \to + \infty } 3 + \mathop {\lim }\limits_{x \to + \infty } \frac{5}{x} = 3 + 0 = 3\).
Vậy \(\mathop {\lim }\limits_{x \to + \infty } \frac{{ – 2x}}{{3 + \frac{5}{x}}} = – \infty \).
g) Ta có: \(\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {4{x^2} + 1} }}{{x + 2}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {{x^2}\left( {4 + \frac{1}{{{x^2}}}} \right)} }}{{x\left( {1 + \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{\left| x \right|\sqrt {4 + \frac{1}{{{x^2}}}} }}{{x\left( {1 + \frac{2}{x}} \right)}}\)
\( = \mathop {\lim }\limits_{x \to – \infty } \frac{{\left( { – x} \right)\sqrt {4 + \frac{1}{{{x^2}}}} }}{{x\left( {1 + \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – \sqrt {4 + \frac{1}{{{x^2}}}} }}{{1 + \frac{2}{x}}}\).
Vì \(\mathop {\lim }\limits_{x \to – \infty } \left( {4 + \frac{1}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to – \infty } 4 + \mathop {\lim }\limits_{x \to – \infty } \frac{1}{{{x^2}}} = 4 + 0 = 4\) nên \(\mathop {\lim }\limits_{x \to – \infty } \sqrt {4 + \frac{1}{{{x^2}}}} = \sqrt 4 = 2\).
Mặt khác, \(\mathop {\lim }\limits_{x \to – \infty } \left( {1 + \frac{2}{x}} \right) = \mathop {\lim }\limits_{x \to – \infty } 1 + \mathop {\lim }\limits_{x \to – \infty } \frac{2}{x} = 1 + 0 = 1\).
Như vậy \(\mathop {\lim }\limits_{x \to – \infty } \frac{{\sqrt {4{x^2} + 1} }}{{x + 2}} = \mathop {\lim }\limits_{x \to – \infty } \frac{{ – \sqrt {4 + \frac{1}{{{x^2}}}} }}{{1 + \frac{2}{x}}} = \frac{{ – 2}}{1} = – 2\).
h) Ta có \(\mathop {\lim }\limits_{x \to 1} \frac{{x – 1}}{{{x^2} – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{x – 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{x + 1}} = \frac{{\mathop {\lim }\limits_{x \to 1} 1}}{{\mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1}} = \frac{1}{{1 + 1}} = \frac{1}{2}\).
i) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} – 5x + 6}}{{x – 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x – 2} \right)\left( {x – 3} \right)}}{{x – 2}} = \mathop {\lim }\limits_{x \to 2} \left( {x – 3} \right) = \mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 3 = 2 + 3 = 5\).
k) \(\mathop {\lim }\limits_{x \to 3} \frac{{ – {x^2} + 4x – 3}}{{{x^2} + 3x – 18}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x – 3} \right)\left( {1 – x} \right)}}{{\left( {x – 3} \right)\left( {x + 6} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{1 – x}}{{x + 6}} = \frac{{\mathop {\lim }\limits_{x \to 3} 1 – \mathop {\lim }\limits_{x \to 3} x}}{{\mathop {\lim }\limits_{x \to 3} x + \mathop {\lim }\limits_{x \to 3} 6}} = \frac{{1 – 3}}{{3 + 6}} = \frac{{ – 2}}{9}\).