Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 30

Show that the statement $\lim \limits_{x \to 3} (x^2 + x - 4) = 8$ 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 $0 < | x - 3 | < \delta$ then $|(x^2 + x - 4 ) - 8 | < \epsilon$

$|(x^2 + x - 4) - 8| < \epsilon \quad \Longrightarrow \quad |x^2 + x - 12| < \epsilon$

To associate $|x^2 + x -12|$ to $|x - 3|$ we can factor and rewrite $|x^2 + x -12|$ to $|(x + 4 )(x - 3)|$ to obtain from the definition

if $0 < | x - 3| < \delta$ then $|(x + 4 )(x - 3)| < \epsilon$

We must find a positive constant $C$ such that $|x + 4 | < C$, so $|x + 4| |x - 3| < C | x - 3|$

From the definition, we obtain

$C | x - 3 | < \epsilon$

$|x - 3| < \frac{\epsilon}{C}$

Again from the definition, we obtain

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

Since we are interested only in values of $x$ that are close to $3$, we assume that $x$ is within a distance $1$ from $3$, that is, $|x - 3| < 1$. Then $2 < x < 4$, so $6 < x + 4 < 8$

Thus, we have $| x + 4 | < 8$ and from there we obtain the value of $C = 8$

But we have two restrictions on $|x - 3|$, namely

$\displaystyle |x - 3|< 1$ and $\displaystyle |x - 3| < \frac{\epsilon}{C} = \frac{\epsilon}{8}$

Therefore, in order for both inequalities to be satisfied, we take $\delta$ to be smaller to $1$ and $\displaystyle \frac{\epsilon}{8}$. The notation for this is $\displaystyle \delta = \text{ min } \left\{1, \frac{\epsilon}{8}\right\}$