Differential and Integral Calculus by Feliciano and Uy: Limit of a Function, Exercise 1.2, Problem 2

Evaluate $\lim\limits_{x\to 3}\left(\frac{4x+2}{x+4}\right)$.

SOLUTION:

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

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

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

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

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

$=\frac{14}{7}$

$=2$