Answer
$\lim\limits_{x\to-5}=-\dfrac{5}{2}.$
Work Step by Step
By Theorem $1.3: \lim\limits_{x\to c}r(x)=r(c)=\dfrac{p(c)}{q(c)}$ where $r(x)=\dfrac{p(x)}{q(x)}.$
$\lim\limits_{x\to-5}\dfrac{5}{x+3}=\dfrac{5}{-5+3}=-\dfrac{5}{2}.$
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