Can you provide a formal proof for:
lim X^2 = 9
x->3
For all $\displaystyle \epsilon>0$ consider the quadratic $\displaystyle d^2 + 6 d - \epsilon$.
By Descartes rule of signs this has exactly one positive root, which I will call $\displaystyle \delta$, which depends only on $\displaystyle \epsilon$ (if you are not happy with Descartes rule of signs here you can find the positive root explicitly if you prefer).
Now suppose $\displaystyle |x-3|< \delta$, then:
$\displaystyle
|x^2-9| = |x-3||x+3| < \delta |x+3|
$.
But:
$\displaystyle |x+3| = |(x-3) + 6| < |x-3| + 6$
so:
$\displaystyle
|x^2-9| = |x-3||x+3| < \delta [|x-3| + 6]< \delta^2 + 6 \delta
$
But by construction $\displaystyle \delta^2 + 6 \delta = \epsilon$.
Hence we have proven that for all $\displaystyle \epsilon>0$ there exists a $\displaystyle \delta>0$ such that $\displaystyle |x-3|< \delta$ implies that $\displaystyle |x^2-9|< \epsilon$, so we have proven that:
$\displaystyle
\lim_{x \to 3} x^2 = 9
$
RonL
(First note that $\displaystyle f(x)=x^2$ is defined on an open interval).
We need to show $\displaystyle |x^2-3|<\epsilon$ for any $\displaystyle \epsilon > 0$ by choosing a $\displaystyle \delta$.
Let use consider $\displaystyle 0<\delta \leq 1$. Then $\displaystyle |x-3|< \delta $ by hypothesis. That means that $\displaystyle |6|+|x-3| < \delta + |6|$ so $\displaystyle |x+3|\leq |6|+|x-3| < \delta + 6$. (Triangle inequality). Hence, $\displaystyle |x^2-9| = |x+3|\cdot |x-3| < (\delta +6)\delta \leq (1+6)\delta = 7\delta$.
So choose,
$\displaystyle \delta = \min \left\{ 1,\frac{\epsilon}{7}\right\}$.