Answer
$\lim\limits_{x\to4}\dfrac{\sqrt{x-3}-1}{x-4}=\dfrac{1}{2}.$
Work Step by Step
$f(x)=\dfrac{\sqrt{x-3}-1}{x-4}=\dfrac{\sqrt{x-3}-1}{x-4}\times\dfrac{\sqrt{x-3}+1}{\sqrt{x-3}+1}$
$=\dfrac{(\sqrt{x-3})^2-(1)^2}{(x-4)(\sqrt{x-3}+1)}=\dfrac{(x-4)}{(x-4)(\sqrt{x-3}+1)}$
$=\dfrac{1}{\sqrt{x-3}+1}=g(x)$
The function $g(x)$ agrees with the function $f(x)$ at all points except $x=4$. Therefore we find the limit as x approaches $4$ of $f(x)$ by substituting the value into $g(x)$.
$\lim\limits_{x\to4}\dfrac{\sqrt{x-3}-1}{x-4}=\lim\limits_{x\to4}\dfrac{1}{\sqrt{x-3}+1}=\dfrac{1}{\sqrt{4-3}+1}=\dfrac{1}{2}.$
