Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 31

Show that the statement $\lim \limits_{x \to -2} (x^2 - 1) = 3$ is correct using the $\epsilon, \delta$ definition of limit

From the definition of the limit
$\text{if } \quad 0 < |x - a| < \delta \quad \text{ then } \quad |f(x) - L| < \varepsilon$

if $0 < | x - (-2) | < \delta $ then $|(x^2 -1 ) -3 | < \epsilon$

$|(x^2 - 1) - 3| < \epsilon \quad \Longrightarrow \quad |x^2 -4| < \epsilon$

To associate $|x^2 -4|$ to $|x + 2|$ we can factor and rewrite $|x^2 -4|$ to $|(x + 2 )(x - 2)|$ to obtain from the definition

if $0 < | x + 2| < \delta$ then $|(x + 2 )(x - 2)| < \epsilon$

We must find a positive constant $C$ such that $|x -2 | < C$, so $|x + 2| |x - 2| < C | x + 2|$

From the definition, we obtain

$C | x +2 | < \epsilon$

$|x+ 2| < \frac{\epsilon}{C}$

Again from the definition, we obtain

$\displaystyle \delta = \frac{\epsilon}{C}$

Since we are interested only in values of $x$ that are close to $-2$, we assume that $x$ is within a distance $1$ from $-2$, that is, $|x + 2| < 1$. Then $-3 < x < -1$, so $-5 < x - 2 < -3$

Thus, we have $| x - 2 | < -3$ and from there we obtain the value of $C = -3$

But we have two restrictions on $|x +2|$, namely

$\displaystyle |x + 2|< 1$ and $\displaystyle |x + 2| < \frac{\epsilon}{C} = \frac{\epsilon}{-3} = -\frac{\varepsilon}{3}$

Therefore, in order for both inequalities to be satisfied, we take $\delta$ to be smaller to $1$ and $\displaystyle -\frac{\varepsilon}{3}$.
The notation for this is $\displaystyle \delta = \text{ min } \left\{1, -\frac{\varepsilon}{3} \right\}$