real analysis - Is this limit proof correct?

I am currently studying the formal defenition of the limit. One of the examples given by my book is the following: Prove that:

$$\lim_{x \to 3} x^2 = 9$$
So, using only the defenition of the limit, I have to prove that for every $\epsilon > 0$ there is a $\delta > 0$ for which the following is true:
$$0 < |x - 3| < \delta \to |x^2 - 9| < \epsilon$$
So after some puzzeling I came up with:
$\delta = \frac{\epsilon}{|x + 3|}$. And I though this was correct: under the assumption that the antecedent is true, we can make the following construct:
$$0 < |x - 3| < \delta \Rightarrow 0 < |x - 3| < \frac{\epsilon}{|x + 3|} \Rightarrow 0 < |x + 3||x - 3| < \epsilon \Rightarrow |x^2 - 9| < \epsilon$$

Q.E.D, I thought. But the book came up with the following solution:
$$\delta = \min{(1,\frac{\epsilon}{7})}$$
Which is a correct solution. So, is mine wrong? Or did the book just provide a different proof?

Answer

Your $\delta$ can't depend on $x$, because $x$ is varying, and so is your $\delta$, but $\delta$ was supposed to be constant. That's the mistake. Let's do a reverse engineering, supposing that $\delta < 1$ (this usually helps).

If $|x-3|<\delta$, then $|x-3|<1 \implies |x| < 3+1 = 4$. And: $$|x^2-9|= |x+3||x-3|\leq (|x|+3)\delta < 7\delta,$$ so that $\delta = \min (1, \epsilon/7)$ will work. So far, just ideas.

Now we check: let $\epsilon > 0$, choose $\delta = \min (1,\epsilon/7)>0$ and take $x \in \Bbb R$ such that $|x-3|<\delta$. Since $\delta < 1$, we also have $|x|<4$. And in these conditions: $$|x^2-9| = |x+3||x-3| \leq (|x|+3)\delta < (4+3)\delta = 7\delta < 7 \frac \epsilon 7 = \epsilon.$$

(You can read more about the idea of supposing that $\delta < 1$ in my answer here.)