# Epsilon-Delta Limit Definition

• Mar 25th 2009, 09:04 AM
Aryth
Epsilon-Delta Limit Definition
Use epsilon-delta proof to show that $\lim_{x\to 4} x^2 = 16$

So far, I have:

Let $\epsilon > 0$

We know that:

$|x^2 - 16| < \epsilon$

and

$|x - 4| < \delta$

Choosing $\delta$:

$|x^2 - 16| < \epsilon$
$|(x + 4)(x - 4)| < \epsilon$
$|x + 4||x - 4| < \epsilon$

From here, I'm having trouble remembering what to do.
• Mar 25th 2009, 12:12 PM
o_O
Arbitrarily assume $\delta \leq 1$ which implies $|x- 4 | < \delta \leq 1$.

But this implies:
$-1 < x - 4 < 1 \ \Leftrightarrow \ 7 < x + 4 < 9 \ \Rightarrow \ |x + 4| < 9$

So we know: $|x+4||x-4| < 9 |x-4| < \epsilon$

Now we can see what we should let $\delta$ be.
• Mar 25th 2009, 01:55 PM
Aryth
I forgot that it was reasonable to insist that x would be within 1 of 4, thanks for the help.
• Mar 25th 2009, 03:20 PM
HallsofIvy
Notice that you must say that $\delta$ is the smaller of $\epsilon/9$ and 1.