Answer
Please see the work step by step below.
Work Step by Step
Given $\epsilon$ $\gt$ 0, we need $\delta$ $\gt$ 0 such that if 0 $\lt$ |$x$ − 2| $\lt$ $\delta$, then | ($x^{2}$ + 2$x$ − 7) − 1 | $\lt$ $\epsilon$ .
| ($x^{2}$ + 2$x$ − 7) − 1 | $\lt$ $\epsilon$ ⇔ | $x^{2}$+ 2$x$ − 8 | $\lt$ $\epsilon$ ⇔ | $x$ + 4 | | $x$ − 2 | $\lt$ $\epsilon$ .
Our goal is to make | $x$ − 2 | small enough so that its product with | $x$ + 4 |
is less than $\epsilon$.
Suppose we first require that | $x$ − 2| $\lt$ 1. Then −1 $\lt$ $x$ − 2 $\lt$ 1 ⇒ 1 $\lt$ $x$ $\lt$ 3 ⇒ 5 $\lt$ $x$ + 4 $\lt$ 7 ⇒ | $x$ + 4 | $\lt$ 7, and this gives us 7 |$x$ − 2| $\lt$ $\epsilon$ ⇒ | $x$ − 2 | $\lt$ $\epsilon$$/$7.
Choose $\delta$ = min {1, $\epsilon$$/$7}.
If 0 $\lt$ |$x$ − 2| $\lt$ $\delta$, then | $x$ − 2 | $\lt$ $\epsilon$$/$7 and |$x$ + 4| $\lt$ 7, so | ( $x^{2}$ + 2$x$ − 7) − 1 | =
|($x$ + 4)($x$ − 2)| = |$x$ + 4| |$x$ − 2| $\lt$ 7($\epsilon$$/$7) = $\epsilon$, as desired.
Thus, $\lim\limits_{x \to 2}$ ($x^{2} + 2$x − 7) = 1 by the definition of a limit.
