Bộ 24 đề thi cuối kì 1 Toán 11 Cánh diều (2023 - 2024) có đáp án - Đề 15

Tìm các giới hạn sau :

31/33

Tìm các giới hạn sau :

            1.\[\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {4x + 1} - 3}}{{x - 2}}\].                                      2.\[\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{{{x^2} + 2023}} + \frac{{x + 4}}{{x + 2}}} \right)\].

0/3000 ký tự
Giải thích

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

2. \[\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{{{x^2} + 2023}} + \frac{{x + 4}}{{x + 2}}} \right)\]\[ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{{{x^2} + 2023}}} \right) + \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x + 4}}{{x + 2}}} \right) = 0 + 2 = 2\]