Thread: Epsilon-Delta Limit Definition

1. Epsilon-Delta Limit Definition

Use epsilon-delta proof to show that $\displaystyle \lim_{x\to 4} x^2 = 16$

So far, I have:

Let $\displaystyle \epsilon > 0$

We know that:

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

and

$\displaystyle |x - 4| < \delta$

Choosing $\displaystyle \delta$:

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

From here, I'm having trouble remembering what to do.

2. Arbitrarily assume $\displaystyle \delta \leq 1$ which implies $\displaystyle |x- 4 | < \delta \leq 1$.

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

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

Now we can see what we should let $\displaystyle \delta$ be.

3. I forgot that it was reasonable to insist that x would be within 1 of 4, thanks for the help.

4. Notice that you must say that $\displaystyle \delta$ is the smaller of $\displaystyle \epsilon/9$ and 1.