Tìm giới hạn của các hàm số sau:

52/55

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

(b) \[\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x - 3} - 2}}{{{x^2} - 49}}\];

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

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

b) \[\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x - 3} - 2}}{{{x^2} - 49}}\]\[ = \mathop {\lim }\limits_{x \to 7} \frac{{x - 7}}{{\left( {{x^2} - 49} \right)\left( {\sqrt {x - 3} + 2} \right)}}\]\[ = \mathop {\lim }\limits_{x \to 7} \frac{1}{{\left( {x + 7} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \frac{1}{{56}}\].

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2\sqrt {5 - x} + \sqrt {7 - 3x} }}{{3{x^2} + 2x - 5}}\)\( = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2 + 4 - 2\sqrt {5 - x} + \sqrt {7 - 3x} - 2}}{{3{x^2} + 2x - 5}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2}}{{3{x^2} + 2x - 5}} + \mathop {\lim }\limits_{x \to 1} \frac{{4 - 2\sqrt {5 - x} }}{{3{x^2} + 2x - 5}} + \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {7 - 3x} - 2}}{{3{x^2} + 2x - 5}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {3x + 5} \right)\left( {\sqrt {x + 3} + 2} \right)}} + 2\mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {3x + 5} \right)\left( {2 + \sqrt {5 - x} } \right)}} - 3\mathop {\lim }\limits_{x \to 1} \frac{1}{{\left( {3x + 5} \right)\left( {\sqrt {7 - 3x} + 2} \right)}} = 0\).