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

Phương pháp giải

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

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

-- Mod Toán 11 HỌC247