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

- anonymous

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- beginnersmind

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 .

- beginnersmind

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?

- zepdrix

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

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## More answers

- anonymous

Yes.

- anonymous

@zepdrix

- anonymous

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

- zepdrix

haha that's weird :D

- zepdrix

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

- anonymous

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

- anonymous

did*

- zepdrix

What a silly billy -_-

- zepdrix

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

- anonymous

Sorry about that...

- zepdrix

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

- zepdrix

Thinkinggg +_+

- anonymous

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

- zepdrix

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 \(\large\rm |x|\lt\sqrt{\epsilon}\)

- zepdrix

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

- anonymous

I got the first blank! x^2

- zepdrix

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

- zepdrix

Do you get multiple guesses or no? :3

- anonymous

I have 100 tries.

- zepdrix

lol

- zepdrix

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

- anonymous

Webassign took it.

- zepdrix

Accepted it? :O Noiceeee

- zepdrix

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.

- zepdrix

Take delta = _______
Hmmmmmmmmmm

- zepdrix

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

- anonymous

It can't understand it.

- zepdrix

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

- anonymous

Yes!

- zepdrix

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

- anonymous

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.

- zepdrix

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

- anonymous

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

- zepdrix

|dw:1442015519645:dw|

- zepdrix

I dunno if you can read my sloppy handwriting there lol
When epsilon puts us 0.2 away from y, how far away from x does that put our delta?

- anonymous

It's better than my chicken scratch. Um... 0.2...?

- anonymous

Nope wait! That was a brain fart!

- zepdrix

So that epsilon puts our line up at 1.2.
That distance between 1.2 and 1 is 0.2, our epsilon.
That line at 1.2 corresponds to an x value of 0.7.
So how far is your 0.7 from 1?

- anonymous

Yeah... I have nothing... This is going to be a long semester with this professor.

- zepdrix

I'm asking you how far it is from 0.7 to 1.
To calculate distance between numbers, you can either count up, or do subtraction.
The distance \(\large\rm \delta\) is going to be 1-0.7.

- zepdrix

That's not our answer though, so don't put that in the box.

- anonymous

I did long before I came here for help.

- zepdrix

??

- zepdrix

I'm just asking a simple question...
what is the distance between 0.7 and 1?
Hello? :o

- anonymous

0.3

- anonymous

I don't know.

- zepdrix

|dw:1442016501759:dw|Ok good, let's call that delta_1, the distance from x on the left side :)

- zepdrix

|dw:1442016540368:dw|Let's check the other side.
If we TAKE AWAY delta from our y, that shows us the corresponding x value of 1.1.
So what is our delta_2? :)

- anonymous

0.1?

- zepdrix

Good!
So we've determined that when we are 0.2 away from our y value,
that corresponds to being 0.3 and 0.1 away from our x value,
\(\large\rm \delta_1=0.3\) and \(\large\rm \delta_2=0.1\)
And \(\large\rm \delta=min\{\delta_1,\delta_2\}\)
So for your delta, choose whichever one is smaller! :)

- anonymous

...This shouldn't have taken me this long...

- zepdrix

lol :)

- anonymous

Thank you for the help!

- zepdrix

yay team \c:/

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