Câu hỏi
Tính \(\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 1} - 2}}{{\sqrt {3x} - 3}}\) bằng?
- A \(\frac{2}{3}.\)
- B \(\frac{1}{3}.\)
- C \(\frac{1}{2}.\)
- D \(1.\)
Phương pháp giải:
- Nhân liên hợp để khử dạng \(\frac{0}{0}\).
Lời giải chi tiết:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 1} - 2}}{{\sqrt {3x} - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{(\sqrt {x + 1} - 2)(\sqrt {x + 1} + 2)(\sqrt {3x} + 3)}}{{(\sqrt {3x} - 3)(\sqrt {3x} + 3)(\sqrt {x + 1} + 2)}} = \mathop {\lim }\limits_{x \to 3} \frac{{(x + 1 - 4)(\sqrt {3x} + 3)}}{{(3x - 9)(\sqrt {x + 1} + 2)}}\\ = \mathop {\lim }\limits_{x \to 3} \frac{{(x - 3)(\sqrt {3x} + 3)}}{{3(x - 3)(\sqrt {x + 1} + 2)}} = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {3x} + 3}}{{3(\sqrt {x + 1} + 2)}} = \frac{{\sqrt {3.3} + 3}}{{3(\sqrt {3 + 1} + 2)}} = \frac{1}{2}\end{array}\)
Chọn: C.