Answer
$$\delta = \varepsilon $$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = 8 \cr
& {\text{By the precise definition of limits}} \cr
& \mathop {\lim }\limits_{x \to a} f\left( x \right) = L \cr
& {\text{if for any number }}\varepsilon {\text{ > 0 there is a corresponding number }}\delta {\text{ > 0}} \cr
& {\text{such that}} \cr
& \left| {f\left( x \right) - L} \right| < \varepsilon {\text{ whenever 0}} < \left| {x - a} \right| < \delta \cr
& {\text{Therefore}} \cr
& L = 8{\text{ and }}f\left( x \right) = \frac{{{x^2} - 16}}{{x - 4}} \cr
& \left| {f\left( x \right) - 8} \right| < \varepsilon \cr
& \left| {\frac{{{x^2} - 16}}{{x - 4}} - 8} \right| < \varepsilon \cr
& {\text{Solving the inequality}} \cr
& - \varepsilon < \frac{{{x^2} - 16}}{{x - 4}} - 8 < \varepsilon \cr
& 8 - \varepsilon < \frac{{{x^2} - 16}}{{x - 4}} < 8 + \varepsilon \cr
& {\text{Factoring}} \cr
& 8 - \varepsilon < \frac{{\left( {x + 4} \right)\left( {x - 4} \right)}}{{x - 4}} < 8 + \varepsilon \cr
& 8 - \varepsilon < x + 4 < 8 + \varepsilon \cr
& 8 - \varepsilon - 8 < x + 4 - 8 < 8 + \varepsilon - 8 \cr
& - \varepsilon < x - 4 < \varepsilon \cr
& \left| {x - 4} \right| < \varepsilon \cr
& {\text{Therefore}} \cr
& \delta = \varepsilon \cr} $$
