## Calculus 10th Edition

$\lim\limits_{x\to1}\dfrac{3x+5}{x+1}=4.$
By Theorem $1.3: \lim\limits_{x\to c}r(x)=r(c)=\dfrac{p(c)}{q(c)}$ where $r(x)=\dfrac{p(x)}{q(x)}.$ $\lim\limits_{x\to1}\dfrac{3x+5}{x+1}=\dfrac{3(1)+5}{1+1}=4.$