Answer
$\lim\limits_{x \to -2} (x^2-1) = 3$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = min\{1, \frac{\epsilon}{5}\}$
Suppose that $\vert x-(-2) \vert \lt \delta$
Then:
$\vert (x^2-1) - 3\vert = \vert x^2-4\vert = \vert (x-2)(x+2)\vert = \vert (x-2)\vert \cdot \vert(x+2)\vert \lt (5)(\delta) \leq (5)(\frac{\epsilon}{5}) = \epsilon$
Therefore, $\lim\limits_{x \to -2} (x^2-1) = 3$
