Bài 3.7 trang 74 SGK Toán 11 tập 1 - Cùng khám phá

Đề bài

Tính các giới hạn sau:

a, $\mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + 3x + 5}}{{x + 1}}$

b, $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x - 6}}{{{x^2} - 4}}$

c, $\mathop {\lim }\limits_{x \to  - 2} \frac{{\sqrt {x + 11}  - 3}}{{x + 2}}$

d, $\mathop {\lim }\limits_{x \to  - \infty } \frac{{3{x^2} + x + 10}}{{2{x^2} - 1}}$

e, $\mathop {\lim }\limits_{x \to  - \infty } \frac{{5{x^3} + 9}}{{{x^4} + 1}}$

g, $\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} + 1} }}{x}$

Lời giải chi tiết

a, Ta có: $\mathop {\lim }\limits_{x \to 0} ({x^2} + 3x + 5) = 5$ và $\mathop {\lim }\limits_{x \to 0} (x + 1) = 1$

Vậy $\mathop {\lim }\limits_{x \to 0} \frac{{{x^2} + 3x + 5}}{{x + 1}} = 5$

b, Ta có : $f(x) = \frac{{{x^2} + x - 6}}{{{x^2} - 4}} = \frac{{(x + 3).(x - 2)}}{{(x - 2).(x + 2)}} = \frac{{x + 3}}{{x + 2}}$

$\mathop {\lim }\limits_{x \to 2} (x + 3) = 5$ và $\mathop {\lim }\limits_{x \to 2} (x + 2) = 4$

Vậy $\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} + x - 6}}{{{x^2} - 4}} = \frac{5}{4}$.

c, Ta có: $f(x) = \frac{{\sqrt {x + 11}  - 3}}{{x + 2}} = \frac{{(\sqrt {x + 11}  - 3)(\sqrt {x + 11}  + 3)}}{{x + 2}} = \frac{{x + 11 - {3^2}}}{{x + 2}} = 1$

$\mathop {\lim }\limits_{x \to  - 2} 1 = 1$

Vậy $\mathop {\lim }\limits_{x \to  - 2} \frac{{\sqrt {x + 11}  - 3}}{{x + 2}} = 1$

d, Ta có: $\mathop {\lim }\limits_{x \to  - \infty } \frac{{3{x^2} + x + 10}}{{2{x^2} - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{3 + \frac{1}{x} + \frac{{10}}{{{x^2}}}}}{{2 - \frac{1}{{{x^2}}}}} = \frac{3}{2}$

e, Ta có: $\mathop {\lim }\limits_{x \to  - \infty } \frac{{5{x^3} + 9}}{{{x^4} + 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{5 + \frac{9}{{{x^4}}}}}{{1 + \frac{1}{{{x^4}}}}} = 5$

g, Ta có: $\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|.\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x.\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to  - \infty } ( - \sqrt {1 + \frac{1}{{{x^2}}}} ) =  - 1$.