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) = x^{2} - 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 = ?

|x^{2} - 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.**