Question

I only have two questions left in my homework that's due tonight before midnight and I need help with them.

1) Prove the statement using the ε, δ definition of a limit.
lim x^2 = 0
x→0
Given ε > 0, we need δ > 0 such that if 0 < |x − 0| <  δ, then |x^2 − 0| <  ε ⇔ (blank) <  ε ⇔ |x|< (blank). Take δ = (blank). Then 0 < |x − 0| <  δ right double arrow implies |x^2 − 0| < ε. Thus, lim x^2 = 0 by the definition of a limit.
x→0

2) Use the given graph of f to find a number δ such that if
|x − 1| < δ then |f(x) − 1| < 0.2 δ = (Blank)
given graph: http://www.webassign.net/scalcet7/2-4-001.gif

Answer 1

For the second one they are asking how far can you go from x = 1 in the x direction so that the values of y are within 0.8 and 1.2 .

Answer 2

E.g, if you go 0.2 to the right, to x = 1.2 is the value of the function still within 0.2 of 1? That is is it in the range [0.8, 1.2]?

What if you only move 0.01? What's the most you can move horizontally to stay in the strip ]0.8, 1.2[ vertically?

Answer 3

So are the (blank)s all the things that you have to fill in?

Answer 4

Yes.

Answer 5

@zepdrix

Answer 6

The notes that my professor gave my class today for these questions don't even come close to them at all.

Answer 7

haha that's weird :D

Answer 8

Oh there is a graph! I didn't even see that part.

Answer 9

I asked him to do the first one I have listed up on the board and he completely does a different question.

Answer 10

did*

Answer 11

What a silly billy -_-

Answer 12

Oh I guess I'm a little confused then.
This looks like two different problems mashed together...

Answer 13

Sorry about that...

Answer 14

Or that's just a second problem? :D
Ok ok ok my bad

Answer 15

Thinkinggg +_+

Answer 16

There, before anyone else joins and gets confused as well.

Answer 17

I think maybe this is what they're looking for...
If x is within delta of zero, \(\large\rm 0\lt |x-0|\lt\delta\)
Then x^2 is within epsilon of the limit value, zero, \(\large\rm |x^2-0|\lt\epsilon\)
Therefore \(\large\rm |x^2|\lt\epsilon\qquad\implies\qquad |x|\cdot|x|\lt\epsilon\)

Since \(\large\rm |x|\lt\delta\) this implies that \(\large\rm |x|\cdot|x|\lt\delta\cdot\delta\)
Choose \(\large\rm \delta\) such that \(\large\rm \delta^2\le\epsilon\)
So that \(\large\rm 0\lt|x-0|\lt\delta\le\sqrt{\epsilon}\)
Which shows that \(\large\rm |x|\lt\sqrt{\epsilon}\)

Answer 18

I've never been very good at these epsilon-delta thingies :\
Lemme try to suggest what they want for the blanks here....

Answer 19

I got the first blank! x^2

Answer 20

So \(\large\rm |x^2-0|\lt\epsilon\) if and only if \(\large\rm x^2\lt\epsilon\)
Ah ok they just simplified :o Hmm

Answer 21

Do you get multiple guesses or no? :3

Answer 22

I have 100 tries.

Answer 23

lol

Answer 24

I'm thinking they want: \(\large\rm |x|\lt\underline{\quad\sqrt{\epsilon}\quad}\)
For the next blank... maybe +_+

Answer 25

Webassign took it.

Answer 26

Accepted it? :O Noiceeee

Answer 27

Maybe what your teacher wanted to do, to get from that step to the next one,\[\large\rm x^2\lt \epsilon\]Is take the square root of both sides of this inequality.\[\large\rm \sqrt{x^2}\lt \sqrt{\epsilon}\]
Recall that \(\large\rm |x|:=\sqrt{x^2}\)
That's what will happen on the left side.
And we shouldn't have any problems with the inequality or square root since epsilon is positive.

Answer 28

Take delta = _______
Hmmmmmmmmmm

Answer 29

Normally we would make a restriction on delta, force it to be small, like delta <= 1,
and then use that to form our epsilon and find a connection between the delta and epsilon.

Then we would let delta be equal to the smaller of those two restrictions.
Like \(\large\rm \delta=min\{\sqrt{\epsilon},1\}\) that or something...
Maybe try it? D: I dunno.. grrr

Answer 30

It can't understand it.

Answer 31

Then maybe just \(\large\rm \delta=\sqrt{\epsilon}\)
:d

Answer 32

Yes!

Answer 33

Is that all of them for number 1? :OO

I'm sorry if that doesn't help you further understand the subject ;c
I know how crazy these can be.
At least we got them filled in though

Answer 34

Yes, indeed! I just want to get this done before 11:59 because that's when my homework is due. Now I just have that one regarding the graph.

Answer 35

Did you read what Beginnersmind wrote about that question? :)

Answer 36

Yes but it confused me further. I'll feel like an idiot if those were the answer.