Câu hỏi
Tính \(\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}}\)bằng?
- A \(\frac{1}{2}.\)
- B \(\frac{9}{8}.\)
- C \(1.\)
- D \(\frac{3}{4}.\)
Phương pháp giải:
- Nhân liên hợp để khử dạng \(\frac{0}{0}\).
Lời giải chi tiết:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}} = \mathop {\lim }\limits_{x \to 2} \frac{{(x - \sqrt {x + 2} )(x + \sqrt {x + 2} )(\sqrt {4x + 1} + 3)}}{{(\sqrt {4x + 1} - 3)(\sqrt {4x + 1} + 3)(x + \sqrt {x + 2} )}} = \mathop {\lim }\limits_{x \to 2} \frac{{({x^2} - x - 2)(\sqrt {4x + 1} + 3)}}{{(4x + 1 - 9)(x + \sqrt {x + 2} )}}\\= \mathop {\lim }\limits_{x \to 2} \frac{{(x + 1)(x - 2)(\sqrt {4x + 1} + 3)}}{{4(x - 2)(x + \sqrt {x + 2} )}} = \mathop {\lim }\limits_{x \to 2} \frac{{(x + 1)(\sqrt {4x + 1} + 3)}}{{4(x + \sqrt {x + 2} )}} = \frac{{(2 + 1)(\sqrt {4.2 + 1} + 3)}}{{4(2 + \sqrt {2 + 2} )}} = \frac{9}{8}\end{array}\)
Chọn: B.