Answer
$\lim\limits_{x\to3}\dfrac{\sqrt{x+6}}{x+2}=\dfrac{3}{5}.$
Work Step by Step
By Theorem $1.3: \lim\limits_{x\to c}r(x)=r(c)=\dfrac{p(c)}{q(c)}$ where $r(x)=\dfrac{p(x)}{q(x)}.$
By Theorem $1.4: \lim\limits_{x\to c}\sqrt[n]{x}=\sqrt[n]{c}.$
$\lim\limits_{x\to3}\dfrac{\sqrt{x+6}}{x+2}=\dfrac{\sqrt{3+6}}{3+2}=\dfrac{3}{5}.$
