Answer
$\delta = min\{2, \frac{\epsilon}{8}\}$ is a possible choice for $\delta$ when we show that: $\lim\limits_{x \to 3} x^2 = 9$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = min\{2, \frac{\epsilon}{8}\}$
Suppose that $\vert x-3 \vert \lt \delta$
Then:
$\vert x^2-9\vert = \vert (x+3)(x-3)\vert = \vert x+3\vert \cdot \vert x-3 \vert \lt (8)(\delta) \leq (8)(\frac{\epsilon}{8}) = \epsilon$
Therefore, $\lim\limits_{x \to 3} x^2 = 9$
