I need to prove that the limit of $x^2$ as $x$ approaches 3 equals 9.
For a given $ > 0$, there exists a $ > 0$ such that:
$$0 < |x - 3| < ⇒ |x² - 9| < $$
Taking the epsilon inequality I started the scratchwork.
$$|x² - 9| = |(x + 3)(x - 3)| < |x + 3| = $$
so we have $ = /|x + 3|$
I know that I have to bound the $|x + 3|$ to find a constant and write $$ in terms of $$, so I let $ = 1$ and:
$$|x - 3| < 1$$
$$\Rightarrow -1 < x - 3 < 1$$
$$\Rightarrow 5 < x + 3 < 7$$
$$\Rightarrow |x + 3| < 7$$
$$ \Rightarrow = /7$$
It seems to be right,but it was told me that I'm not setting $ = 1$, but $ < 1$. Why? My point here is not "why 1 and not any other number" but why the inequality sign and not the equal sign? Letting $ = 1$ wouldn't still be useful because $|x - 3|$ is still less than 1?
I think you're misinterpreting what was said to you. You can't be setting $\delta = 1$ because you're ending up with $\delta = \epsilon/7$, and that would contradict your setting of $\delta=1$. So then what do you even mean when you say that you have set $\delta$?
What that chain of calculations is saying is that "When $\delta=1$ does not work, then $|x-3|<1$, and so then $\delta=\epsilon/7$ should work." That's why the answer is the minimum of $1$ and $\epsilon/7$. The value of $\delta=1$ would not work when $\epsilon=0.01$ and the value of $\delta=\epsilon/7$ would not work when $\epsilon=1000$.