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