Limit of $x^2$ as $x$ approaches $3$. so I have a few question(s) about limits. I have to prove that $\lim_{x \to 3} x^2 = 9$ using the $\epsilon-\delta$ definition of limit. This question is from James Stewart's Calculus book. Now the book says: If you let $\delta$ be the smaller of the numbers 1 and $\frac{\epsilon}{7}$, show that this $\delta$ works.
How did it even get 1? I mean I've searched online multiple times and people have just guessed let $\delta = 1$. Why, how? Please explain. So far my working has been this:
Suppose we have an arbitrary $\epsilon > 0$ given. We want to find a $\delta > 0$ such at $0 < |x-3| < \delta \implies |x^{2} - 9| < \epsilon$.
We start from $|x^{2} - 9| < \epsilon$
$|x-3||x+3| < \epsilon$
The book says show that if $|x-3| < 1$, then $|x+3| < 7$.
$|x-3| < 1$ gives us $-1 < x-3 < 1$, after adding 6 to all sides we get, $-5 < x-3 < 7$, and so $|x-3| < 7$.
We have $|x-3||x+3| < \epsilon$, since $|x+3| < 7$, we have $|x-3||x+3| < 7|x-3|$.
We need a $\delta$ such that $7|x-3| < \epsilon$ or $|x-3| < \frac{\epsilon}{7}$. So we choose $\delta = \frac{\epsilon}{7}$ and do the whole reverse working, etc. First of all, is the working uptil this part correct? And, how do we get $\delta = 1$? The book also says choose the minimum of 1 and $\frac{\epsilon}{7}$ and show that the delta works. But how do we get delta = 1?
 A: We fixed $\epsilon > 0$ and our goal is to find $\delta > 0$ such that
$$
\lvert x^2 - 9 \rvert = \lvert x-3 \rvert \lvert x+3 \rvert < \epsilon
$$
whenever $\lvert x-3 \rvert < \delta$.
For now, suppose we are given $\delta > 0$. We try to see how we can bound
$$
\lvert x^2 - 9 \rvert = \lvert x-3 \rvert \lvert x+3 \rvert.
$$
assuming that $\lvert x - 3 \rvert  < \delta$. Clearly, we have
$$
\lvert x^2 - 9 \rvert = \lvert x-3 \rvert \lvert x+3 \rvert < \delta \lvert x+3 \rvert.
$$
We now wish to bound $\lvert x + 3\rvert$. Since we are only given a bound on $\lvert x-3\rvert$, we use the triangle inequality as follows:
$$
\lvert x+3 \rvert = \lvert (x-3) + 6 \rvert \leq \lvert x-3 \rvert +6 < \delta + 6.
$$
Combining our inequalities, we see that
$$
\lvert x^2 - 9 \rvert = \lvert x-3 \rvert \lvert x+3 \rvert < \delta \left( \delta + 6\right).
$$
In order to avoid dealing with a quadratic, we would like to bound $\delta + 6$ by a constant number. $\color{blue}{\text{If we know that }\delta \leq 1, \text{then it must be that } \delta + 6 \leq 7.}$ Our problem is thus simplified. In particular, we have
$$
\lvert x^2 - 9 \rvert = \lvert x-3 \rvert \lvert x+3 \rvert < \delta \left( \delta + 6\right) \leq 7\delta.
$$
Finally, we ask when $7\delta < \epsilon$. This is the case whenever
$$
\delta < \frac{\epsilon}{7}.
$$
In conclusion, if $\delta \leq 1$ and $\delta  \leq \epsilon/7$ then
$$
\lvert x^2 - 9 \rvert = \lvert x-3 \rvert \lvert x+3 \rvert < \delta \left( \delta + 6\right) \leq 7\delta < \epsilon
$$
whenever $\lvert x-3 \rvert <\delta$.
Hence, for any $\epsilon >0$ we pick $\delta > 0$ such that
$$
\delta \leq 1 \quad\text{and}\quad \delta \leq \frac{\epsilon}{7}.
$$
The choice of $\delta = \min\{1, \epsilon/7\}$ is an example of such a number.
