Solution of A Limit Computed from the Epsilon-Delta Definition

Let $\varepsilon>0$ and put

$$\delta=\min\left\{1,\ \frac{\varepsilon}{7}\right\} ,$$

which is positive because $0<1$ and $0<\varepsilon/7$.

Let $x\in\mathbb{R}$ satisfy $0<|x-3|<\delta$.

Step 1: bounding $|x+3|$. Since $\delta\le 1$ we have $|x-3|<1$, so $-1<x-3<1$ and hence $2<x<4$. Adding $3$ gives $5<x+3<7$. In particular $0<x+3$, so $|x+3|=x+3<7$.

Step 2: factoring. By the field arithmetic of Elementary Arithmetic in an Ordered Field,

$$x^2-9=(x-3)(x+3) ,$$

and by the multiplicativity of the absolute value recorded in Properties of the Absolute Value in an Ordered Field,

$$|x^2-9|=|x-3|\,|x+3| .$$

Step 3: the estimate. Since $0\le|x-3|$ and $|x+3|<7$, we have $|x-3|\,|x+3|\le|x-3|\cdot 7$, using the order arithmetic of Elementary Order Arithmetic in an Ordered Field. Since $|x-3|<\delta\le\varepsilon/7$ and $0<7$,

$$|x-3|\cdot 7<\frac{\varepsilon}{7}\cdot 7=\varepsilon .$$

Combining the last two displays with Step 2 gives $|x^2-9|<\varepsilon$.

For each $\varepsilon>0$ we have thus exhibited a $\delta>0$ with the required property, so the real number $9$ satisfies the condition of Limit of a Real Function at a Point of an Interval (i) for $f$ at the point $3$ of the interval $\mathbb{R}$. By Uniqueness of the Limit of a Real Function at a Point of an Interval (ii) no other real number has that property, so the limit is well defined and we may write

$$\lim_{x\to 3}f(x)=9 .$$