lim_{x->2} \dfrac{ \sqrt{4x+1}-3}{x^2-4} 01/11/2021 Bởi Savannah lim_{x->2} \dfrac{ \sqrt{4x+1}-3}{x^2-4}
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` Bình luận
$\lim\limits_{x\to 2}\dfrac{ \sqrt{4x+1}-3}{x^2-4}$ $=\lim\limits_{x\to 2}\dfrac{4x+1-3^2}{(x^2-4)(\sqrt{4x+1}+3)}$ $=\lim\limits_{x\to 2}\dfrac{4(x-2)}{(x-2)(x+2)(\sqrt{4x+1}+3)}$ $=\lim\limits_{x\to 2}\dfrac{4}{(x+2)(\sqrt{4x+1}+3)}$ $=\dfrac{4}{(2+2).(\sqrt{4.2+1}+3)}$ $=\dfrac{1}{6}$ Bình luận
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`
$\lim\limits_{x\to 2}\dfrac{ \sqrt{4x+1}-3}{x^2-4}$
$=\lim\limits_{x\to 2}\dfrac{4x+1-3^2}{(x^2-4)(\sqrt{4x+1}+3)}$
$=\lim\limits_{x\to 2}\dfrac{4(x-2)}{(x-2)(x+2)(\sqrt{4x+1}+3)}$
$=\lim\limits_{x\to 2}\dfrac{4}{(x+2)(\sqrt{4x+1}+3)}$
$=\dfrac{4}{(2+2).(\sqrt{4.2+1}+3)}$
$=\dfrac{1}{6}$