Đáp án:
$\lim\limits_{x\to 2^+}\dfrac{x-2+\sqrt{x-2}}{\sqrt{x -2}} =1$
Giải thích các bước giải:
$\quad \lim\limits_{x\to 2^+}\dfrac{x-2+\sqrt{x-2}}{\sqrt{x -2}}$
$= \lim\limits_{x\to 2^+}\dfrac{\sqrt{x-2}(\sqrt{x-2} +1)}{\sqrt{x-2}}$
$=\lim\limits_{x\to 2^+}(\sqrt{x-2} +1)$
$=\sqrt{2-2} +1$
$=1$