Limits questions

- Jamierox4ev3r

Limits questions

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- Jamierox4ev3r

Given that \[\lim_{x \rightarrow 2}(5x-5) =5\], find values of \(\delta\) that correspond to \(\epsilon\)=0.1, \(\epsilon\)=0.05, and \(\epsilon\)=0.01

- ganeshie8

finally you have started reviewing epsilon delta!

- Jamierox4ev3r

Looks like it :P

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## More answers

- Jamierox4ev3r

I would think you would start off by making this equation:
-0.1<5x-5<0.1

- Jamierox4ev3r

then solving for x
-0.1<5x-5<0.1
+5 +5 +5
-------------
4.9<5x<5.1
/5 /5 /5
---------
0.98

- Jamierox4ev3r

I'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

0.98

- Jamierox4ev3r

right

- ganeshie8

so \(\delta = 0.02\) works

- Jamierox4ev3r

so you don't need to subtract from whatever the value of a is? (which in this problem, it's 2)

- ganeshie8

what that means is
if you're with in \(0.02\) around \(x=2\),
the value of function stays within \(0.1\) around \(5\)

- ganeshie8

a graph might help
you want to spend some time and see whats going on

- ganeshie8

or you may continue with rest of the problems and hope things become clear after doing few problems

- Jamierox4ev3r

It's funny, I've already done a lot of problems. And things seemed pretty straightforward until i got to this one.

- Jamierox4ev3r

For example, I had a problem that read, "Find a number \(\delta\) such that if \[\left| x-2 \right|=\delta\], then \(\left| 5x-10\right|=\epsilon\), where \(\epsilon =0.1\)

- ganeshie8

ikr! 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

^ agreed. But this problem that preceded...I was able to get it with no problem. my process looked like this:
-0.1<5x-10<0.1
+10 +10 +10
------------------
9.9<5x<10.1
/5 /5 /5
----------
1.98

- Jamierox4ev3r

from 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

So 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

let me ask you a question there

- Jamierox4ev3r

Sure

- ganeshie8

you have
```
1.98

- Jamierox4ev3r

yes

- ganeshie8

that is wrong
how can both give you 0.02 ?
double check

- Jamierox4ev3r

wait...

- Jamierox4ev3r

2.02-2=0.02
2-1.98=0.02
^^
they both do give 0.02 o-o

- ganeshie8

Oops sry, you're right!

- ganeshie8

so it seems the same \(\delta\) works in both cases

- Jamierox4ev3r

yeah :P weird how the solutions are exactly the same, regardless of the functions being different

- Jamierox4ev3r

Why is that? do you know?

- ganeshie8

they both are kinda same functions, they only differ by a constant

- ganeshie8

f(x) = 5x-5
g(x) = 5x-10

- Jamierox4ev3r

that's true...I suppose that makes sense, thy only vary by where in the graph they start

- Jamierox4ev3r

Another thing i don't get is that with the
5x-10, you use 2 in order to find \(\delta\)
but with
5x-5, you use 1 in order to find \(\delta\)
I don't know why I get hung up on these tiny details XD

- ganeshie8

teamviewer ?

- Jamierox4ev3r

o-o pardon?

- ganeshie8

you don't have teamviewer/skype?

- Jamierox4ev3r

nope

- Jamierox4ev3r

I could screenshot you what I'm looking at o-o

- ganeshie8

not necessary, i have a khan academy video that explains exactly this
one sec..

- ganeshie8

here is it is
https://www.khanacademy.org/math/differential-calculus/limits_topic/epsilon_delta/v/epsilon-delta-limit-definition-1

- Jamierox4ev3r

omg @ganeshie8 that's actually a super clear explanation, I got it! Thank you u.u

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