# Homework Help: Illustrate the definition of a limit

1. Sep 9, 2011

### fillipeano

1. The problem statement, all variables and given/known data
For the limit

lim
x → 2 (x^3 − 3x + 5) = 7

illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.)

2. Relevant equations

3. The attempt at a solution
I've gotten x^3-3x=2.2, but I'm lost after that. My teacher gave a terrible description on what to do, and the book is of no help whatsoever.

2. Sep 9, 2011

### lanedance

so lets start by writing x=2+d, with \delta =|d| and substitute in
|((2+d)^3 − 3(2+d) + 5-7| <0.2

3. Sep 9, 2011

### fillipeano

Alright, so now I've got
d^3 + 6d^2 + 9 < 0.2

4. Sep 9, 2011

### lanedance

That doesn't look quite right, i would check your steps. Also you can't drop the absolute value sign

5. Sep 9, 2011

### fillipeano

|(2+d)^3 -6 - 3d - 2| < 0.2
|d^3 +6d^2 + 12d + 8 - 6 - 3d - 2| < 0.2

= |d^3 + 6d^2 + 9d| < 0.2

6. Sep 9, 2011

### Tomer

I don't see at all how this is helpful, and cannot recall when this serves as a good approach.

fillipeano-
I don't see how it is easier to solve it for a specific $\epsilon$ rather than a general one, so let's try and guide you on the general one. Then we'll just substitute the values given.

So let $\epsilon$ > 0.
As usually, you need to start from the end.
Let's say f(x) = x3 -3x + 5
I can understand of course how you got to the equation you wrote, but it should actually be an inequality. We need to find a $\delta$, so that for each x for which |x-2|<$\delta$| holds, |f(x) - 7|<$\epsilon$ also holds. (instead of saying $\epsilon$ you could say "0.2")

So, let see how we can "control" the expression |f(x)-7|:

|x3 -3x + 5 - 7| = |x3 - 3x-2|.
Now we're stuck.
However, we know that this has to do somehow with "|x-2|", because we want to use that fact that |x-2|<$\delta$|.
That should make you think about polynomial division. Do you know the technique?
Then you could have |x3 - 3x-2| = |x-2| * |P(x)| < $\delta$ |P(x)|,
where P(x) is some polynomial of second order which you might also be able to control. :)
See if you could take it from there...

7. Sep 9, 2011

### lanedance

in this case it simplifies down to
|d^3 + 6d^2 + 9d| = |d^2 + 6d+ 9||d| = (d+3)^2|d|<e

though it is the same thing really whether you work with |x-2| or |d|, just preference

8. Sep 9, 2011

### Tomer

But what is the justification of this substitution? One needs to show that |f(x) - 7| < $\epsilon$. So why do you try manipulating |f(x+$\delta$) - 7|?

9. Sep 9, 2011

### HallsofIvy

Because the whole problem is to find $|\delta$ so that $f(2+ \delta)- 7|< \epsilon$. It is NOT "to show that $|f(x)- 7|<\epsilon$" because that, in general, is not true.

10. Sep 9, 2011

### Tomer

Of course, but wouldn't you then want to show that |$f(2- \delta)- 7|< \epsilon$ as well?
Looks like a strange approach to me, but if everyone's approving... :-)
(for the same suitable $\delta$ that is)