Show $\lim_{x \to 4} \frac{x^2 - 16}{x-4} = 8$

Question

My proof so far is as follows:

Let $\epsilon > 0$ be given. Choose a $\delta =$ (not sure) such that $0 < |x-4| < \delta$.

We have $\left|\frac{x^2-16}{x-4} - 8\right| = \left|\frac{x^2 - 8x + 16}{x-4}\right| = \left|x-4\right|$

I am not sure how to show that $$\left|\frac{x^2-16}{x-4} - 8\right| = \left|\frac{x^2 - 8x + 16}{x-4}\right| = \left|x-4\right|< \epsilon$$

Answer 1

$\lim\limits_{x \to 4} {x^2 - 16 \over x-4} = \lim\limits_{x \to 4}\ (x+4) = 8$.

Answer 2

Let $$\delta = \epsilon$$

$$0< |x-4|< \delta \implies $$

$$|\frac{x^2-16}{x-4} - 8| = |\frac{x^2 - 8x + 16}{x-4}| = |x-4|< \epsilon.$$