Answer
$\lim\limits_{x\to3}\dfrac{\sqrt{x+1}-2}{x-3}=\dfrac{1}{4}.$
Work Step by Step
$f(x)=\dfrac{\sqrt{x+1}-2}{x-3}=\dfrac{\sqrt{x+1}-2}{x-3}\times\dfrac{\sqrt{x+1}+2}{\sqrt{x+1}+2}$
$=\dfrac{(\sqrt{x+1})^2-(2)^2}{(x-3)(\sqrt{x+1}+2)}=\dfrac{(x-3)}{(x-3)(\sqrt{x+1}+2)}$
$=\dfrac{1}{\sqrt{x+1}+2}=g(x).$
The function $g(x)$ agrees with the function $f(x)$ at all points except $x=3$. Therefore, we find the limit as x approaches $3$ of $f(x)$ by substituting the value into $g(x)$.
$\lim\limits_{x\to3}\dfrac{\sqrt{x+1}-2}{x-3}=\lim\limits_{x\to3}\dfrac{1}{\sqrt{x+1}+2}=\dfrac{1}{\sqrt{3+1}+2}=\dfrac{1}{4}.$
