Tính $\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}}$ bằng?
Trả lời bởi giáo viên
$\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {x + 2} }}{{\sqrt {4x + 1} - 3}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - \sqrt {x + 2} )(x + \sqrt {x + 2} )(\sqrt {4x + 1} + 3)}}{{(\sqrt {4x + 1} - 3)(\sqrt {4x + 1} + 3)(x + \sqrt {x + 2} )}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{({x^2} - x - 2)(\sqrt {4x + 1} + 3)}}{{(4x + 1 - 9)(x + \sqrt {x + 2} )}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 1)(x - 2)(\sqrt {4x + 1} + 3)}}{{4(x - 2)(x + \sqrt {x + 2} )}} \\= \mathop {\lim }\limits_{x \to 2} \dfrac{{(x + 1)(\sqrt {4x + 1} + 3)}}{{4(x + \sqrt {x + 2} )}} \\= \dfrac{{(2 + 1)(\sqrt {4.2 + 1} + 3)}}{{4(2 + \sqrt {2 + 2} )}} = \dfrac{9}{8}\end{array}$
Hướng dẫn giải:
- Nhân liên hợp để khử dạng $\dfrac{0}{0}$.