Illustrate the definition of a limit

[fillipeano]

Homework Statement

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.)

Homework Equations

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.

 

[lanedance]

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

 

[fillipeano]

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

 

[lanedance]

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

 

[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

 

[Tomer]

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

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...

 

[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

 

[Tomer]

in this case it simplifies down to
|d^3 + 6d^2 + 9d| = |d^2 + 6d+ 9||d| = (d+3)^2|d|<0.2

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

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|?

 

[HallsofIvy]

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|?

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

 

[Tomer]

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

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