Answer
Please see below.
Work Step by Step
We want to prove that $$\lim_{x \to c}(mx+b)=mc+b,$$where $m \neq 0$, by using $\epsilon - \delta$ definition; that is, we must show that for each $\epsilon >0$, there exists a $\delta >0$ such that $|(mx+b)-(mc+b)|< \epsilon$ whenever $|x-c|< \delta$.
Now, we have $|(mx+b)-(mc+b)|=|m||x-c|$. So, by choosing $\delta =\frac{\epsilon }{|m|}$ ($\delta$ is well-defined since $m \neq 0$) we conclude that$$|x-c|< \delta =\frac{\epsilon }{|m|} \quad \Rightarrow \quad |(mx+b)-(mc+b)|=|m||x-c|$$.
