Answer
$\lim\limits_{x \to 2} x^2 = 8$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = min\{1, \frac{\epsilon}{19}\}$
Suppose that $\vert x-2 \vert \lt \delta$
Then:
$\vert x^2-8\vert = \vert (x^2+2x+4)(x-2)\vert = \vert (x^2+2x+4)\vert \cdot \vert(x-2)\vert \lt (19)(\delta) \leq (19)(\frac{\epsilon}{19}) = \epsilon$
Therefore, $\lim\limits_{x \to 2} x^2 = 8$
