How do I find the delta analytically for $f(x)$ with a degree other than $1$ So I know how to find the $\delta$ in $f(x)$ that can be factored into a degree of $1$, or I can solve it by solving $L + \epsilon = f(x) $ then finding the distance from $a$. But how do I find the delta analytically for an $f(x)$ for example a degree $2$.
Example: $f(x) = x^2$; $a =3$; $L = 9$; $\epsilon = 0.5$
$$ 0 < |x+3|<\delta $$
$$ |x^2-9| < 0.5 $$
$$ |x+3| |x-3| < 0.5$$
I get stuck at this part when solving the epsilon side.
$$ |x+3| |x-3| < 0.5$$
I can't factor $|x+3|$ like in functions of degree $1$. Can someone please explain?

If $0\le \delta \le 1$, then $|x-3|\lt \delta$, then $x+3\lt 7$. Thus if furthermore $|x-3|\lt \frac{\epsilon}{7}$, then $|x^2-9|\lt \epsilon$.
So if we pick $\delta=\min(1,\epsilon/7)$, then $|x^2-9|\lt \epsilon$ whenever $|x-3|\lt \delta$.
Here we did not find the cheapest $\delta$ that works.  Proving that the limit exists does not require that we find the cheapest $\delta$, given $\epsilon$. It only requires that we show there is a $\delta$.
Remark: If you really want to find the cheapest $\delta$, for say $\epsilon=0.5$, you need to solve the inequality $|x^2-9|\lt 0.5$. Separate this inequality into two cases (i) |x|\ge 3$ and (ii) |x|\lt 3$. for (i), we are looking at the inequality $x^2\lt 9.5$, and for (ii) we are looking at $x^2\gt 8.5$. Now use the calculator.
