Could I get a critique of this epsilon-delta limit proof?

Question

$\lim \limits_{x \to 3}$ ${(x^2-2)}$ = 7

So I want to find some $\delta$ > 0 such that for every $\epsilon$ > 0:

$\lvert x^2-9\rvert$ < $\epsilon$ $\iff$ 0 < $\lvert x-3\rvert$ < $\delta$

By a property of absolute value:

$\lvert x+3\rvert$$\lvert x-3\rvert$ < $\epsilon$ $\iff$ 0 < $\lvert x-3\rvert$ < $\delta$

So the left inequality can be written as:

$\lvert x-3\rvert$ < ${\epsilon\over \lvert x+3\rvert}$

Since we're dealing with values of x that are near 3, we can arbitrarily say that we wish to concern ourselves with x in the range:

2 < x < 4

Therefore the maximum value of $\lvert x+3\rvert$ is 7. Since this produces the minimum value of ${\epsilon\over \lvert x+3\rvert}$, we wish to set this as our delta, as it provides the strictest stipulation on the inequality.

$\delta$ = ${\epsilon\over 7}$

Therefore, on the right side of the biconditional statement:

$\lvert x-3\rvert$ < ${\epsilon\over 7}$

We can multiply ${\epsilon\over 7}$ by 7 to get $\epsilon$ on the right side; since the value of $\lvert x+3\rvert$ will never be greater in seven and will also never be negative, the inequality will still hold if we multiply the left side by it.

$\lvert x+3\rvert$$\lvert x-3\rvert$ < $\epsilon$

Simplifying that inequality, we have shown that:

$\lvert x^2-9\rvert$ < $\epsilon$ $\iff$ 0 < $\lvert x-3\rvert$ < $\delta$

I realize that this is most likely frightful. I'm very new to writing proofs of any kind, but it's something I really would like to learn all I can about. Thanks for your time if you choose to respond.

Edit (updated proof):

I want to find some $\delta$ > 0 such that, for some value of $\epsilon$ > 0:

0 < $\lvert x-3\rvert$ < $\delta$ $\Rightarrow$ $\lvert x^2-9\rvert$ < $\epsilon$

Rewriting the right inequality:

$\lvert x-3\rvert$ < ${\epsilon\over \lvert x+3\rvert}$

Restricting x such that:

2 < x < 4

means that the minimum value that ${\epsilon\over \lvert x+3\rvert}$ can attain is ${\epsilon\over 7}$; therefore, the maximum value that $\lvert x-3\rvert$ can obtain is ${\epsilon\over 7}$.

Therefore, setting $\delta$ = ${\epsilon\over 7}$:

$\lvert x-3\rvert$ < ${\epsilon\over 7}$

Multiplying the left side by $\lvert x+3\rvert$ while multiplying the right by 7 is acceptable because $\lvert x+3\rvert$ will always be less than 7, thus preserving the inequality. That leaves us with:

0 < $\lvert x-3\rvert$ < $\delta$ $\Rightarrow$ $\lvert x^2-9\rvert$ < $\epsilon$

Answer 1

You pretty much have all the right pieces. Note that we're not dealing with a biconditional though; we want to show that: $$ 0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon $$ Here's a cleaned up version of your proof.

---

Given any $\epsilon > 0$, consider $\delta = \min\{1, \epsilon/7\} > 0$. Then observe that if $0 < |x - 3| < \delta$, then: \begin{align*} |x^2 - 9| &= |x - 3||x + 3| \\ &< \frac{\epsilon}{7}|x + 3| &\text{since }|x - 3| < \delta \leq \frac{\epsilon}{7} \\ &= \frac{\epsilon}{7}|(x - 3) + (6)| \\ &\leq \frac{\epsilon}{7}\left(|x - 3| + |6|\right) &\text{by the triangle inequality} \\ &< \frac{\epsilon}{7}\left(1 + |6|\right) &\text{since }|x - 3| < \delta \leq 1 \\ &= \epsilon \end{align*} as desired. $~~\blacksquare$