\begin{align*}

\lim_{x\to 3}\left(\frac{4x+2}{x+4}\right)& =\frac{\lim\limits_{x\to 3}\left(4x+2\right)}{\lim\limits_{x\to 3}\left(x+4\right)}\\

& =\frac{\lim\limits_{x\to 3}\left(4x\right)+\lim\limits_{x\to 3}\left(2\right)}{\lim\limits_{x\to 3}\left(x\right)+\lim\limits_{x\to 3}\left(4\right)}\\

& =\frac{4\cdot \lim\limits_{x\to 3}\left(x\right)+2}{3+4}\\

& =\frac{4\cdot 3+2}{3+4}\\

& =\frac{12+2}{7}\\

& =\frac{14}{7}\\

& =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\

\end{align*}