Tính giới hạn lim x → 1 √ x + 3 + √ 2x + 7 − 5 /(2x − 2) kết quả làm tròn đến hàng phần trăm.

Trả lời: 0,29

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

\[ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 3} \right) - 4}}{{\left( {2x - 2} \right)\left( {\sqrt {x + 3} + 2} \right)}} + \mathop {\lim }\limits_{x \to 1} \frac{{\left( {2x + 7} \right) - 9}}{{\left( {2x - 2} \right)\left( {\sqrt {2x + 7} + 3} \right)}}\]

\[ = \mathop {\lim }\limits_{x \to 1} \frac{1}{{2\left( {\sqrt {x + 3} + 2} \right)}} + \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {2x + 7} + 3}}\]\[ = \frac{1}{8} + \frac{1}{6}\]\[ = \frac{7}{{24}} \approx 0,29\].