Giải bài 10 trang 85 sách bài tập toán 11 - Chân trời sáng tạo tập 1
Tính các giới hạn sau: a) lim; b) \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} + 2{x^2}}}{{3{x^2} + 1}}; c) \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2} - 2x + 3} .
Đề bài
Tính các giới hạn sau:
a) \mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + 2{x^2} - 1} \right);
b) \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} + 2{x^2}}}{{3{x^2} + 1}};
c) \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2} - 2x + 3} .
Phương pháp giải - Xem chi tiết
Sử dụng kiến thức về quy tắc tính giới hạn vô cực để tính:
a) Nếu \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = L > 0,\mathop {\lim }\limits_{x \to - \infty } g\left( x \right) = - \infty thì \mathop {\lim }\limits_{x \to - \infty } \left[ {f\left( x \right)g\left( x \right)} \right] = - \infty
b) Nếu \mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = L > 0,\mathop {\lim }\limits_{x \to + \infty } g\left( x \right) = + \infty thì \mathop {\lim }\limits_{x \to + \infty } \left[ {f\left( x \right)g\left( x \right)} \right] = + \infty
c) Nếu \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = L > 0,\mathop {\lim }\limits_{x \to - \infty } g\left( x \right) = + \infty thì \mathop {\lim }\limits_{x \to - \infty } \left[ {f\left( x \right)g\left( x \right)} \right] = + \infty
Lời giải chi tiết
a) \mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + 2{x^2} - 1} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {{x^3}\left( {1 + \frac{2}{x} - \frac{1}{{{x^3}}}} \right)} \right]
Vì \mathop {\lim }\limits_{x \to - \infty } {x^3} = - \infty ;\mathop {\lim }\limits_{x \to - \infty } \left( {1 + \frac{2}{x} - \frac{1}{{{x^3}}}} \right) = \mathop {\lim }\limits_{x \to - \infty } 1 + \mathop {\lim }\limits_{x \to - \infty } \frac{2}{x} - \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{{x^3}}} = 1 > 0
Do đó, \mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + 2{x^2} - 1} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {{x^3}\left( {1 + \frac{2}{x} - \frac{1}{{{x^3}}}} \right)} \right] = - \infty
b) \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} + 2{x^2}}}{{3{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\frac{{1 + \frac{2}{x}}}{{3 + \frac{1}{{{x^2}}}}}} \right]
Ta có: \mathop {\lim }\limits_{x \to + \infty } x = + \infty ,\mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{2}{x}}}{{3 + \frac{1}{{{x^2}}}}} = \frac{{1 + \mathop {\lim }\limits_{x \to + \infty } \frac{2}{x}}}{{3 + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{{x^2}}}}} = \frac{1}{3} > 0
Do đó, \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^3} + 2{x^2}}}{{3{x^2} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\frac{{1 + \frac{2}{x}}}{{3 + \frac{1}{{{x^2}}}}}} \right] = + \infty
c) \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2} - 2x + 3} \mathop {\lim }\limits_{x \to - \infty } \left[ {\left| x \right|\sqrt {1 - \frac{2}{x} + \frac{3}{{{x^2}}}} } \right] = \mathop {\lim }\limits_{x \to - \infty } \left[ { - x\sqrt {1 - \frac{2}{x} + \frac{3}{{{x^2}}}} } \right]
Ta có: \mathop {\lim }\limits_{x \to - \infty } \left( { - x} \right) = + \infty ;\mathop {\lim }\limits_{x \to - \infty } \sqrt {1 - \frac{2}{x} + \frac{3}{{{x^2}}}} = \sqrt {1 - \mathop {\lim }\limits_{x \to - \infty } \frac{2}{x} + \mathop {\lim }\limits_{x \to - \infty } \frac{3}{{{x^2}}}} = 1 > 0
Do đó, \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2} - 2x + 3} = \mathop {\lim }\limits_{x \to - \infty } \left[ { - x\sqrt {1 - \frac{2}{x} + \frac{3}{{{x^2}}}} } \right] = + \infty