Tìm các giới hạn sau:

12/13

a) \(\mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{x + 5}}\);

b) \(\mathop {\lim }\limits_{x \to 6} \frac{{\sqrt {x - 2} - 2}}{{{x^2} - 6x}}.\)

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

a) Ta có \(\mathop {\lim }\limits_{x \to - 5} \frac{{2{x^2} + 9x - 5}}{{x + 5}} = \mathop {\lim }\limits_{x \to - 5} \frac{{\left( {2x - 1} \right)\left( {x + 5} \right)}}{{x + 5}} = \mathop {\lim }\limits_{x \to - 5} \left( {2x - 1} \right) = 2 \cdot \left( { - 5} \right) - 1 = - 11\).

b) Ta có \(\mathop {\lim }\limits_{x \to 6} \frac{{\sqrt {x - 2} - 2}}{{{x^2} - 6x}} = \mathop {\lim }\limits_{x \to 6} \frac{{\left( {\sqrt {x - 2} - 2} \right)\left( {\sqrt {x - 2} + 2} \right)}}{{x\left( {x - 6} \right)\left( {\sqrt {x - 2} + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to 6} \frac{{\left( {x - 2} \right) - 4}}{{x\left( {x - 6} \right)\left( {\sqrt {x - 2} + 2} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 6} \frac{{x - 6}}{{x\left( {x - 6} \right)\left( {\sqrt {x - 2} + 2} \right)}}\)\( = \mathop {\lim }\limits_{x \to 6} \frac{1}{{x\left( {\sqrt {x - 2} + 2} \right)}}\)\( = \frac{1}{{6 \cdot \left( {\sqrt {6 - 2} + 2} \right)}} = \frac{1}{{6 \cdot \left( {2 + 2} \right)}} = \frac{1}{{24}}\).