Jump to content


Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account.
As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends.

Of course, you can also enjoy our collection of amazing optical illusions and cool math games.

If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top.
If you have a website, we would appreciate a little link to BrainDen.

Thanks and enjoy the Den :-)
Guest Message by DevFuse
 

Photo
- - - - -


  • Please log in to reply
2 replies to this topic

#1 harvey45

harvey45

    Senior Member

  • Members
  • PipPipPipPip
  • 629 posts

Posted 20 August 2010 - 11:46 PM

I am having a bit of trouble understanding how to work with the formal definition of limits. Here is the problem I'm stuck on:

f(x) = x2 - 1
Find δ such that if 0 < |x - 2| < δ then |f(x) - 3| < 0.2

I have this so far:

L = 3
ε = 0.2
c = 2
δ = ?

|x2 - 4| < 0.2
0 < x - 2 < δ

So yeah, all I've really gotten is the values plugged in and simplified. Could someone give me an explanation on how to finish this?

**for reference, the formal def. of a limit is:
"Lim(x->c) f(x) = L" means that for each ε > 0 there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε
  • 0

#2 archlordbr

archlordbr

    Advanced Member

  • Members
  • PipPipPip
  • 242 posts

Posted 21 August 2010 - 12:30 AM

Maybe you could replace the < with = on both equations, then calculate the 2 possible values for your function, and out of those you get the values for x, but one of them will be smaller than 2, so you get the other one and find your delta. I'm not sure this would be mathematically correct though. I don't know why, but I'm studying limits at college right now and the teacher never mentioned anything about finding the delta. He only said that the limit exists if there's a delta even for a really small ε. Maybe it's because it is an Engineering class.
  • 0

#3 mmiguel1

mmiguel1

    Advanced Member

  • Members
  • PipPipPip
  • 392 posts

Posted 26 August 2010 - 03:36 AM

I am having a bit of trouble understanding how to work with the formal definition of limits. Here is the problem I'm stuck on:

f(x) = x2 - 1
Find δ such that if 0 < |x - 2| < δ then |f(x) - 3| < 0.2

I have this so far:

L = 3
ε = 0.2
c = 2
δ = ?

|x2 - 4| < 0.2
0 < x - 2 < δ

So yeah, all I've really gotten is the values plugged in and simplified. Could someone give me an explanation on how to finish this?

**for reference, the formal def. of a limit is:
"Lim(x->c) f(x) = L" means that for each ε > 0 there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε



You are proving that the limit of x^2-1 as x approaches 2 is 3.
This means that if you pick any ε which is greater than zero:
it is possible to find a δ, also greater than zero, such that
whenever x is between 2 + δ and 2 - δ, then f(x) will be between 3+ε and 3-ε.

To say that the limit exists means that it is possible to find such a δ, the limit does not signify what the values of δ or ε are, only that it is POSSIBLE to find a δ given an ε as described above.

If it is not possible to find a δ given an ε, then the limit does not exist at x=2.
If a limit does exist at x=2, then it will have a unique value which is 3 in this case.
The limit cannot have more than one value (i.e. 3 in this case) where it is possible to find a δ given an ε.

Hopefully that illuminates the definition a little.

So, for your particular problem, the first step is to assume you have selected a value for ε.
ε=0.2 or something, it doesn't matter, and we'll just call whatever value you pick ε.

To prove that the limit exists as you stated, you must show that it is possible to find a δ for any ε you pick such that the constraints in the definition of the limit are satisfied.

So for ε=0.2, is it possible to find such a δ?
|x^2-4| < 0.2
can be re-expressed as
-0.2 < x^2-4 < 0.2
and
3.8 < x^2 < 4.2
and since x^2 is always an increasing function in this range (i.e. if y > x, then y^2 > x^2 and vice versa)
then equivalently,
sqrt(3.8) < x < sqrt(4.2)

We want to show that a δ can be found that ensures that for all x in the range 0 < |x-2| < δ, the constraint:
sqrt(3.8) < x < sqrt(4.2)
is always satisfied.


0 < |x-2| < δ which means that:
x does not equal 2
and
-δ < x-2 < δ

Keeping in mind that x does not equal 2, we simplify the second expression:
2-δ < x < 2+δ

Now, compare:
sqrt(3.8) < x < sqrt(4.2)
to
2-δ < x < 2+δ

To ensure that sqrt(3.8) < x < sqrt(4.2) is satisfied, we only need to make sure that
2-δ >= sqrt(3.8) and 2+δ <= sqrt(4.2)
This means that
δ <= 2-sqrt(3.8) and δ <= sqrt(4.2)-2
and we must not forget that by definition, δ > 0

As long as 2-sqrt(3.8) is greater than zero AND sqrt(4.2)-2 is greater than zero, then a δ can be found.
Using a calculator, 2-sqrt(3.8) = 0.050641131
sqrt(4.2)-2=0.049390153

The minimum of the two is the important one.

So basically, assuming ε=0.2 means that
any δ where 0 < δ < 0.049390153 will satisfy the constraints in the definition of the limit for L=3.

Let's generalize this a bit, let's say we never fixed ε to be 0.2, but rather left it as ε.
In this case, we can follow the same steps, except whenever we used 0.2, just use ε instead.

You get something like
-0.2 < x^2-4 < 0.2
becomes
-ε < x^2-4 < ε
and
sqrt(3.8) < x < sqrt(4.2)
becomes
sqrt(4-ε) < x < sqrt(4 + ε)

and
0 < δ < min(0.050641131,0.049390153)
which is the same as
0 < δ < min(2-sqrt(3.8),sqrt(4.2)-2)
becomes
0 < δ < min(2-sqrt(4-ε),sqrt(4+ε)-2)

So for what values of ε is it possible to find a δ?
It is possible to find a δ as long as 2-sqrt(4-ε) > 0 and sqrt(4+ε)-2 > 0
This means
sqrt(4-ε) < 2
4-ε < 4
4 < 4 + ε
0 < ε
and
4+ε > 4
or
ε > 0

It is therefore possible to find a δ for any ε>0.

This completes the proof that the limit exists at this point.

Edited by mmiguel1, 26 August 2010 - 03:45 AM.

  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users