Tính lim từ x đến 2 x-căn bậc hai (x+2)/căn bậc hai (4x+1)-3 bằng? A.1/2 B.9/8 C.1 D.3/4

* Hướng dẫn giải

$\underset{x\to 2}{\lim }\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}$

$=\underset{x\to 2}{\lim }\frac{(x-\sqrt{x+2})(x+\sqrt{x+2})(\sqrt{4x+1}+3)}{(\sqrt{4x+1}-3)(\sqrt{4x+1}+3)(x+\sqrt{x+2})}$

$\begin{array}{l}=\underset{x\to 2}{\lim }\frac{(x^2-x-2)(\sqrt{4x+1}+3)}{(4x+1-9)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x+1)(x-2)(\sqrt{4x+1}+3)}{4(x-2)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}\\ =\frac{(2+1)(\sqrt{4.2+1}+3)}{4(2+\sqrt{2+2})}=\frac{9}{8}\end{array}$