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anonymous
 one year ago
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/scalcet7/24001.gif
anonymous
 one year ago
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/scalcet7/24001.gif

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beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.0E.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
 one year ago
Best ResponseYou've already chosen the best response.1So are the (blank)s all the things that you have to fill in?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The notes that my professor gave my class today for these questions don't even come close to them at all.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh there is a graph! I didn't even see that part.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I asked him to do the first one I have listed up on the board and he completely does a different question.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1What a silly billy _

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh I guess I'm a little confused then. This looks like two different problems mashed together...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Or that's just a second problem? :D Ok ok ok my bad

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There, before anyone else joins and gets confused as well.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I think maybe this is what they're looking for... If x is within delta of zero, \(\large\rm 0\lt x0\lt\delta\) Then x^2 is within epsilon of the limit value, zero, \(\large\rm x^20\lt\epsilon\) Therefore \(\large\rm x^2\lt\epsilon\qquad\implies\qquad x\cdotx\lt\epsilon\) Since \(\large\rm x\lt\delta\) this implies that \(\large\rm x\cdotx\lt\delta\cdot\delta\) Choose \(\large\rm \delta\) such that \(\large\rm \delta^2\le\epsilon\) So that \(\large\rm 0\ltx0\lt\delta\le\sqrt{\epsilon}\) Which shows that \(\large\rm x\lt\sqrt{\epsilon}\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I've never been very good at these epsilondelta thingies :\ Lemme try to suggest what they want for the blanks here....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got the first blank! x^2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So \(\large\rm x^20\lt\epsilon\) if and only if \(\large\rm x^2\lt\epsilon\) Ah ok they just simplified :o Hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Do you get multiple guesses or no? :3

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I'm thinking they want: \(\large\rm x\lt\underline{\quad\sqrt{\epsilon}\quad}\) For the next blank... maybe +_+

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Accepted it? :O Noiceeee

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.1Take delta = _______ Hmmmmmmmmmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Normally 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
 one year ago
Best ResponseYou've already chosen the best response.0It can't understand it.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Then maybe just \(\large\rm \delta=\sqrt{\epsilon}\) :d

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Is 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1Did you read what Beginnersmind wrote about that question? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes but it confused me further. I'll feel like an idiot if those were the answer.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0It's better than my chicken scratch. Um... 0.2...?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope wait! That was a brain fart!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah... I have nothing... This is going to be a long semester with this professor.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I'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 10.7.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1That's not our answer though, so don't put that in the box.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did long before I came here for help.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I'm just asking a simple question... what is the distance between 0.7 and 1? Hello? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442016501759:dwOk good, let's call that delta_1, the distance from x on the left side :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442016540368:dwLet'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? :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good! 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
 one year ago
Best ResponseYou've already chosen the best response.0...This shouldn't have taken me this long...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for the help!
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