Đáp án:
Giải thích các bước giải:
$\lim _{x\to 2}\left(\dfrac{\:\sqrt{4x+1}-3}{x^2-4}\right)$
$=\lim _{x\to \:2}\left(\dfrac{\dfrac{4x-8}{\sqrt{4x+1}+3}}{x^2-4}\right)$
$=\lim _{x\to \:2}\left(\dfrac{\dfrac{4x-8}{\sqrt{4x+1}+3}}{x^2-4}\right)$
$=\dfrac{4}{\left(\sqrt{4\cdot \:2+1}+3\right)\left(2+2\right)}$
`=1/6`