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

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

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