Đáp án:
$\lim\limits_{x\to 2}\dfrac{\sqrt{x+2} + \sqrt{x+7} -5}{x-2}=\dfrac{5}{12}$
Giải thích các bước giải:
$\quad \lim\limits_{x\to 2}\dfrac{\sqrt{x+2} + \sqrt{x+7} -5}{x-2}$
$= \lim\limits_{x\to 2}\dfrac{\sqrt{x+2} - 2}{x-2} + \lim\limits_{x\to 2}\dfrac{\sqrt{x+7} -3}{x-2}$
$= \lim\limits_{x\to 2}\dfrac{\left(\sqrt{x+2} - 2\right)\left(\sqrt{x+2} + 2\right)}{(x-2)\left(\sqrt{x+2} +2\right)} + \lim\limits_{x\to 2}\dfrac{\left(\sqrt{x+7} -3\right)\left(\sqrt{x+7} +3\right)}{(x-2)\left(\sqrt{x+7} +3\right)}$
$= \lim\limits_{x\to 2}\dfrac{1}{\sqrt{x+2} +2} + \lim\limits_{x\to 2}\dfrac{1}{\sqrt{x+7} + 3}$
$= \dfrac{1}{\sqrt{2+2} +2} +\dfrac{1}{\sqrt{2+7} +3}$
$= \dfrac{5}{12}$
