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Jamierox4ev3r
 one year ago
Limits questions
Jamierox4ev3r
 one year ago
Limits questions

This Question is Closed

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1Given that \[\lim_{x \rightarrow 2}(5x5) =5\], find values of \(\delta\) that correspond to \(\epsilon\)=0.1, \(\epsilon\)=0.05, and \(\epsilon\)=0.01

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3finally you have started reviewing epsilon delta!

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1Looks like it :P

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1I would think you would start off by making this equation: 0.1<5x5<0.1

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1then solving for x 0.1<5x5<0.1 +5 +5 +5  4.9<5x<5.1 /5 /5 /5  0.98<x<1.02

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1I'm just kind of confused as to what you would do here... Since a=2, wouldn't you subtract 0.98 and 1.02 from 2?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.30.98<x<1.02 thats same as : x < 0.02 yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3so \(\delta = 0.02\) works

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1so you don't need to subtract from whatever the value of a is? (which in this problem, it's 2)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3what that means is if you're with in \(0.02\) around \(x=2\), the value of function stays within \(0.1\) around \(5\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3a graph might help you want to spend some time and see whats going on

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3or you may continue with rest of the problems and hope things become clear after doing few problems

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1It's funny, I've already done a lot of problems. And things seemed pretty straightforward until i got to this one.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1For example, I had a problem that read, "Find a number \(\delta\) such that if \[\left x2 \right=\delta\], then \(\left 5x10\right=\epsilon\), where \(\epsilon =0.1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3ikr! often times the concepts look so easy when you read them but you only get to really learn them only after actually solving the problems

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1^ agreed. But this problem that preceded...I was able to get it with no problem. my process looked like this: 0.1<5x10<0.1 +10 +10 +10  9.9<5x<10.1 /5 /5 /5  1.98<x<2.002

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1from there, I subtracted 2 from 2.02, and 1.98 from two, which both gave me 0.02. So I concluded that \(\delta =0.02\), which was a correct answer according to the computer software I'm using.

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1So I assumed from then that I knew what I was doing XD I suppose I was just confused about the fact that the functions are different, but they wield the same answers

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3let me ask you a question there

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you have ``` 1.98<x<2.002 ``` then you say : ``` from there, I subtracted 2 from 2.02, and 1.98 from two, which both gave me 0.02.`

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that is wrong how can both give you 0.02 ? double check

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.12.022=0.02 21.98=0.02 ^^ they both do give 0.02 oo

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Oops sry, you're right!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3so it seems the same \(\delta\) works in both cases

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1yeah :P weird how the solutions are exactly the same, regardless of the functions being different

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1Why is that? do you know?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3they both are kinda same functions, they only differ by a constant

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3f(x) = 5x5 g(x) = 5x10

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1that's true...I suppose that makes sense, thy only vary by where in the graph they start

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1Another thing i don't get is that with the 5x10, you use 2 in order to find \(\delta\) but with 5x5, you use 1 in order to find \(\delta\) I don't know why I get hung up on these tiny details XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you don't have teamviewer/skype?

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1I could screenshot you what I'm looking at oo

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3not necessary, i have a khan academy video that explains exactly this one sec..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3here is it is https://www.khanacademy.org/math/differentialcalculus/limits_topic/epsilon_delta/v/epsilondeltalimitdefinition1

Jamierox4ev3r
 one year ago
Best ResponseYou've already chosen the best response.1omg @ganeshie8 that's actually a super clear explanation, I got it! Thank you u.u
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