# Finding Delta Given Epsilon with a Quadratic Function

1. Jul 13, 2017

### Cosmophile

• Thread moved from the technical forums, so no Homework Help Template is shown
Hey, everyone! I'm helping a friend through his calculus course and we've come across something that has stumped me (see: the title). When I learned calculus, our treatment of the epsilon-delta definition of the limit was, at best, brief. Anyway, here is the problem:

Given $\lim_{x \rightarrow 3} (x^2 -2) = 7$ and $\epsilon = 0.2$,
find $\delta$ such that for all $x, \quad$ $0<|x-3|<\delta \implies |x^2 - 9| < 0.2$.

Here is my work so far:

$$0 < |x-3| < \delta \quad, \quad |x^2 - 9| < 0.2$$
$$0< |x-3| < \delta \quad, \quad |x+3||x-3| < 0.2$$
$$0 < |x-3| <\delta \quad, \quad |x-3| < \frac{0.2}{|x+3|}$$

...And there is where I get lost. I've found a connection between the absolute value expressions, but I can't find any way to make progress from there. I'm sure this is the wrong approach for any problems of this form where $f(x)$ has a degree $>1$, so any help is greatly appreciated!

2. Jul 13, 2017

### scottdave

I would not do this: |x+3||x-3|, you really only need the absolute value of the whole thing. But I would look at cases.
So for the case when x² ≥ 9, then |x² - 9| < 0.2 is the same as just (x² - 9) < 0.2
Then for the case when x² < 9, then |x² - 9| is the same as (x² - 9) > -0.2

That should make it easier than having two factors with absolute values, and dividing by absolute values, etc.

Last edited: Jul 14, 2017
3. Jul 13, 2017

### Staff: Mentor

4. Jul 13, 2017

### Staff: Mentor

5. Jul 14, 2017

### scottdave

Thanks, I fixed it.

6. Jul 14, 2017

### SammyS

Staff Emeritus
Good.

By the way: Congrats on becoming an official Homework Helper !

Last edited: Jul 14, 2017
7. Jul 15, 2017

### scottdave

Thank You.
Thank You. I'm honored.

8. Jul 15, 2017

### SammyS

Staff Emeritus
The method suggested by @scottdave can get you the maximum value that can be used for $\ \delta \,$, but that's not the usual method used in a proof for the case where ε has an arbitrary value.

You have $\displaystyle \ |x+3||x-3| < 0.2\,,\$ giving $\displaystyle \ |x-3| < \frac{0.2}{|x+3|}\ .\$ So, if you can find some "upper bound" on $\ | x+3|\,,\$ then you can find a value for $\ \delta\,.$

Certainly, we must have $\ \delta < 1\ .$ That implies that x is between 2 and 4 and thus $\ | x+3|\$ must be between 5 and 7.

Use this information to find a value for $\ \delta\,.$