Answer
$\lim\limits_{x\to4^-}\dfrac{\sqrt{x}-2}{x-4}=\dfrac{1}{4}.$
Work Step by Step
$\lim\limits_{x\to4^-}\dfrac{\sqrt{x}-2}{x-4}=\lim\limits_{x\to4^-}\dfrac{(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\lim\limits_{x\to4^-}\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{\sqrt{4}+2}=\dfrac{1}{4}.$
