Answer
$\lim\limits_{x \to 4} \frac{x^2-2x-8}{x-4} = 6$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = \epsilon$
Suppose that $\vert x-4 \vert \lt \delta$.
Then: $\vert \frac{x^2-2x-8}{x-4}-6 \vert = \vert \frac{(x-4)(x+2)}{x-4}-6 \vert = \vert (x+2)-6 \vert = \vert x-4 \vert \lt \delta = \epsilon$
Therefore, $\lim\limits_{x \to 4} \frac{x^2-2x-8}{x-4} = 6$
