Answer
$$\delta = \frac{\varepsilon }{2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 3} \left( { - 2x + 8} \right) = 2 \cr
& {\text{By the precise definition of limits}} \cr
& \mathop {\lim }\limits_{x \to a} f\left( x \right) = L \cr
& {\text{if for any number }}\varepsilon {\text{ > 0 there is a corresponding number }}\delta {\text{ > 0}} \cr
& {\text{such that}} \cr
& \left| {f\left( x \right) - L} \right| < \varepsilon {\text{ whenever 0}} < \left| {x - a} \right| < \delta \cr
& {\text{Therefore}} \cr
& L = 2{\text{ and }}f\left( x \right) = - 2x + 8 \cr
& \left| {f\left( x \right) - 2} \right| < \varepsilon \cr
& \left| { - 2x + 8 - 2} \right| < \varepsilon \cr
& \left| { - 2x + 6} \right| < \varepsilon \cr
& {\text{Solving the inequation}} \cr
& - \varepsilon < - 2x + 6 < \varepsilon \cr
& - \varepsilon - 6 < - 2x < \varepsilon - 6 \cr
& \frac{{ - \varepsilon - 6}}{{ - 2}} > x > \frac{{\varepsilon - 6}}{{ - 2}} \cr
& \frac{\varepsilon }{2} + 3 > x > 3 - \frac{\varepsilon }{2} \cr
& 3 - \frac{\varepsilon }{2} < x < \frac{\varepsilon }{2} + 3 \cr
& - \frac{\varepsilon }{2} < x - 3 < \frac{\varepsilon }{2} \cr
& \left| {x - 3} \right| < \frac{\varepsilon }{2} \cr
& {\text{Therefore}} \cr
& \delta = \frac{\varepsilon }{2} \cr} $$
