BrainDen.com - Brain Teasers
• 0

## Question

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

## 2 answers to this question

• 0

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.

##### Share on other sites
• 0

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

## Create an account

Register a new account