Bài 1 trang 79 SGK Toán 11 tập 1 - Chân trời sáng tạo

Đề bài

Tìm các giới hạn sau:

a) $\mathop {\lim }\limits_{x \to  - 2} \left( {{x^2} - 7x + 4} \right)$  

b) $\mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{{x^2} - 9}}$                                    

c) $\mathop {\lim }\limits_{x \to 1} \frac{{3 - \sqrt {x + 8} }}{{x - 1}}$

Lời giải chi tiết

a) $\mathop {\lim }\limits_{x \to  - 2} \left( {{x^2} - 7x + 4} \right) = \mathop {\lim }\limits_{x \to  - 2} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to  - 2} \left( {7x} \right) + \mathop {\lim }\limits_{x \to  - 2} 4$

                                                $= \mathop {\lim }\limits_{x \to  - 2} \left( {{x^2}} \right) - 7\mathop {\lim }\limits_{x \to  - 2} x + \mathop {\lim }\limits_{x \to  - 2} 4 = {\left( { - 2} \right)^2} - 7.\left( { - 2} \right) + 4 = 22$

b) $\mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{1}{{x + 3}} = \frac{{\mathop {\lim }\limits_{x \to 3} 1}}{{\mathop {\lim }\limits_{x \to 3} x + \mathop {\lim }\limits_{x \to 3} 3}} = \frac{1}{{3 + 3}} = \frac{1}{6}$

c) $\mathop {\lim }\limits_{x \to 1} \frac{{3 - \sqrt {x + 8} }}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {3 - \sqrt {x + 8} } \right)\left( {3 + \sqrt {x + 8} } \right)}}{{\left( {x - 1} \right)\left( {3 + \sqrt {x + 8} } \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{3^2} - \left( {x + 8} \right)}}{{\left( {x - 1} \right)\left( {3 + \sqrt {x + 8} } \right)}}$

                                         $= \mathop {\lim }\limits_{x \to 1} \frac{{1 - x}}{{\left( {x - 1} \right)\left( {3 + \sqrt {x + 8} } \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {3 + \sqrt {x + 8} } \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - 1}}{{3 + \sqrt {x + 8} }}$

                                         $= \frac{{\mathop {\lim }\limits_{x \to 1} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 1} 3 + \mathop {\lim }\limits_{x \to 1} \sqrt {x + 8} }} = \frac{{ - 1}}{{3 + \sqrt {1 + 8} }} =  - \frac{1}{6}$