Question

Little limits help here?

Answer 1

No idea how to do this.

Answer 2

What are those symbols?

Answer 3

Delta

Answer 4

The other one is epsilon

Answer 5

Okay first calculate |x - a| and |f(x) - L|

Answer 6

Which gives us |x-1| and |-3x^2+x-2|

Answer 7

make that +2

Answer 8

Understand so far

Answer 9

-0.5 < -3x^2+x+2 < 0.5
and
-d < x-1 < d

Answer 10

You are going to have to play around with these inequalities a bit.

Answer 11

i'm curious why is it |-3x^2+x-2|
and not |-3x^3+x-2|
?

Answer 12

|-3x^3+x+2|*

Answer 13

we could have 1-d < x < 1+d, then plug plug these values into the above inequality to find bounds for d.

Answer 14

that is, plug in 1+d for x and 1-d for x in the -0.5 < -3x^2+x+2 < 0.5 inequality

Answer 15

BTW, have you noted that \(2+x-x^{3} = (x-1)(3x^{2}+3x+2)\)?

Answer 16

Oh, so that's why.

Answer 17

Does corresponding delta mean any eligible delta, or the largest delta, though?

Where did you get 1.05 and 0.95?

Answer 18

I think he got those values by adding increasing the one from lim x-->1
by 0.05 and decreasing by 0.05.

Answer 19

The requirement is AT MOST. Thus, the greater must be selected.

The problem statement includes \(\epsilon = 0.05\).

Answer 20

Look, you could always pick something like 10^(-100000000) if you wanted an easy out.

Answer 21

I always get those backwards!!

What we need is if \(0 < |x - c| < \delta\), then \(|f(x) - L| < \epsilon\).

Answer 22

Yep.

Answer 23

We can start with -0.5 < 2+x-3x^2 < 0.5
And factor it to -0.5 < -(x-1) (3 x^2+3 x+2) < 0.5
mean while we know -d < x - 1 < d

Answer 24

In English, how close do we have to stay in the neighborhood of 1, in order to ensure that the function stays in the the .05-neighborhood of 7. Using the right solution, this time, we have 0.01006 and 0.00994, so this time pick the lesser. Then we wait for wio to do it the right (more rigorous) way.

Answer 25

If you don't want to be rigorous, pick delta = 10^(-10000) and then show that for that it never exceeds the bounds.

The rigorous ways gives you actual bounds though

Answer 26

Either way rigor goes into showing the function never exceeds a point on the interval. You cant just assume the endpoints are max and mins

Are you going to use calculus to show it has certain bounds? Just overlook the limits done to carry out the calc?

Answer 27

-0.5/(3 x^2+3 x+2) < x-1 < 0.5/(3 x^2+3 x+2)

Answer 28

|x-1| < d
|x-1| < 0.5/(3 x^2+3 x+2)


We know that 3 x^2+3 x+2 can be as large as infinity since it is an upward parabola. We also know its smallest point would be at its vertex.

Answer 29

Right, not a complaint. (except that you keep using 0.5 instead of 0.05) Just waiting to see your approach.

Given \(0 < |x-1| < \delta\), we require \(|2+x-3x^{3}|<0.05\)

Or \(|-(x-1)(3x^{2}+3x+2)| = |(x-1)(3x^{2}+3x+2)| < 0.05\)

Then, since the minimum value of \(3x^{3}+3x+2\) is 5/4 (over all of \(\mathbb{R}\)), we have:

\(|(x-1)| < \dfrac{0.05}{(3x^{2}+3x+2)} \le \dfrac{0.05}{(5/4)} = 0.04\)

Answer 30

It would not be terrible to pick some smaller interval, smaller than \(\mathbb{R}\).

Answer 31

Vertex form of 3 x^2+3 x+2 is 3 (x+1/2)^2+5/4, which means 5/4 is the smallest that it can get.

Answer 32

0.05 / (5/4) = 0.2 / 5 = 0.04

Answer 33

would want to double check it

Answer 34

I've been keeping up so far, but I'll need more than a few mins to process and truly understand it.. thanks guys~

Answer 35

Another thing you could do is find all points where

f(x) = 6.95
f(x) = 7.05

Then pick the one which is closest to x=1.

Answer 36

Since it would be the first one to actually exceed our boundary.

Answer 37

You would need to find roots of a cubic though, to do that method.

Answer 38

Alright, thanks wio~

Answer 39

Hmmm, I might have made an error in my prior reasoning to get 0.04

Answer 40

because the function is not linear, you're gonna have two values of delta of which the smaller one is chosen

Answer 41

^ You're late xD

Answer 42

:DD