Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 33

Show that the another possible value for $\delta$ that would satisfy the $\lim \limits_{x \to 3} x^2 = 9$ is $\delta = \text{ min } \left\{ 2, \frac{\varepsilon}{8}\right\}$


From the definition of the limit

if $0 < | x - 3 | < \delta$ then $|(x^2 ) -9 | < \varepsilon$

To associate $|x^2 - 9|$ to $|x - 3|$ we can factor and rewrite $|x^2 -9|$ to $|(x + 3 )(x - 3)|$ to obtain from the definition

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

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

From the definition, we obtain

$C | x - 3 | < \varepsilon$

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

Again from the definition, we obtain

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

Since we are interested only in values of $x$ that are close to $3$, we assume that $x$ is within a distance $2$ from $3$, that is, $|x - 3| < 2$. Then $1 < x < 5$, so $x+3 < 8$

Therefore, we have $| x + 3| < 8$ and from there we obtain the value of $C = 8$

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

$\displaystyle |x - 3|< 2$ and $\displaystyle |x - 3| < \frac{\varepsilon}{C} = \frac{\varepsilon}{8}$

In order for both inequalities to be satisfied, we take $\delta$ to be smaller to $2$ and $\displaystyle \frac{\varepsilon}{8}$. The notation for this is $\displaystyle \delta = \text{ min } \left\{2, \frac{\varepsilon}{8}\right\}$