Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 32

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

To associate $|x^3 -8|$ to $|x - 2|$ we can factor and rewrite $|x^3 -8|$ to $|(x - 2 )(x^2 +2x + 4)|$ to obtain from the definition

if $0 < | x - 2| < \delta$ then $|(x - 2 )(x^2 +2x + 4)| < \epsilon$

We must find a positive constant $C$ such that $|x^2 +2x + 4 | < C$, so $|x^2 +2x + 4| |x - 2| < C | x - 2|$

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 $1 < x < 3$, so $x^2 + 2x + 4 < (3)^2 + 2(3) + 4 = 19$

Thus, we have $| x^2 + 2x + 4 | < 19$ and from there we obtain the value of $C = 19$

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

$\displaystyle |x - 2|< 4$ and $\displaystyle |x - 2| < \frac{\epsilon}{C} = \frac{\epsilon}{19}$

Therefore, in order for both inequalities to be satisfied, we take $\delta$ to be smaller to $1$ and $\displaystyle \frac{\epsilon}{19}$. The notation for this is $\displaystyle \delta = \text{ min } \left\{1, \frac{\epsilon}{19}\right\}$