Answer
$\lim\limits_{x \to 2} (x^2+2x-7) = 1$
Work Step by Step
Let $\epsilon \gt 0$ be given.
Let $\delta = min\{1, \frac{\epsilon}{7}\}$
Suppose that $\vert x-2 \vert \lt \delta$
Then:
$\vert (x^2+2x-7) - 1\vert = \vert x^2+2x-8\vert = \vert (x+4)(x-2)\vert = \vert (x+4)\vert \cdot \vert(x-2)\vert \lt (7)(\delta) \leq (7)(\frac{\epsilon}{7}) = \epsilon$
Therefore, $\lim\limits_{x \to 2} (x^2+2x-7) = 1$
