# Math Help - Delta-Epsilon proof

1. ## Delta-Epsilon proof

Can you provide a formal proof for:

lim X^2 = 9
x->3

2. Originally Posted by CrazyAsian
Can you provide a formal proof for:

lim X^2 = 9
x->3
For all $\epsilon>0$ consider the quadratic $d^2 + 6 d - \epsilon$.

By Descartes rule of signs this has exactly one positive root, which I will call $\delta$, which depends only on $\epsilon$ (if you are not happy with Descartes rule of signs here you can find the positive root explicitly if you prefer).

Now suppose $|x-3|< \delta$, then:

$
|x^2-9| = |x-3||x+3| < \delta |x+3|
$
.

But:

$|x+3| = |(x-3) + 6| < |x-3| + 6$

so:

$
|x^2-9| = |x-3||x+3| < \delta [|x-3| + 6]< \delta^2 + 6 \delta
$

But by construction $\delta^2 + 6 \delta = \epsilon$.

Hence we have proven that for all $\epsilon>0$ there exists a $\delta>0$ such that $|x-3|< \delta$ implies that $|x^2-9|< \epsilon$, so we have proven that:

$
\lim_{x \to 3} x^2 = 9
$

RonL

3. Wow...

I mean, yeah... Just what I was thinking.

4. Originally Posted by CrazyAsian
Can you provide a formal proof for:

lim X^2 = 9
x->3
(First note that $f(x)=x^2$ is defined on an open interval).

We need to show $|x^2-3|<\epsilon$ for any $\epsilon > 0$ by choosing a $\delta$.

Let use consider $0<\delta \leq 1$. Then $|x-3|< \delta$ by hypothesis. That means that $|6|+|x-3| < \delta + |6|$ so $|x+3|\leq |6|+|x-3| < \delta + 6$. (Triangle inequality). Hence, $|x^2-9| = |x+3|\cdot |x-3| < (\delta +6)\delta \leq (1+6)\delta = 7\delta$.
So choose,
$\delta = \min \left\{ 1,\frac{\epsilon}{7}\right\}$.