Prove the statement using the epsilon-delta definition of limit: lim (x?0) x4 = 0

Question

Prove the statement using the epsilon-delta definition of limit: lim (x→0) x⁴ = 0

 

 

I have no idea how to do this and the book has no answers to practice problems.

 

#19, practice test

Answer 1

Hi Mia,

 

It can be fairly tedious to prove a limit using ε-δ, but this should give you the idea. 

 

The ε-δ definition of limit goes something like this:

 

Assuming you are trying to prove lim_(x->a)f(x) = L: 

 

If x is in the open interval (a - δ, a + δ) and x ≠ a, then f(x) is in the open interval (L - ε, L + ε).

 

It's kind of like a game. If you tell me how close you want f(x) to be to L (ε), then I will tell you how close x has to be to a (δ) to make that happen.

 

If f(x) is in the open interval shown above, another way to state that is:

 

| f(x) - L | < ε, i.e. the distance from f(x) to L is less than some given number ε.

 

We usually start with that statement and work towards 0 < | x - a | < δ.

 

In your problem a = 0 and L = 0. Also f(x) = x⁴.

 

| x⁴ - 0 | < ε

 

If we take the 4th root of both sides, we get:

 

|x| < ε^1/4

 

So, that is our choice of delta.

 

If you give me how close you want f(x) to be to 0 (like within .0016), then all I need to do is choose an x value that is within the 4th root of .0016 (.2). So guaranteed for this function if x = .18, then x⁴ will be within .0016 of the limit 0. 

 

After having determined that we are going to let δ = ε^1/4, the proof goes like this:

 

0 < | x - 0 | <  ε^1/4

 

| x⁴| <  ε

 

| x⁴ - 0| < ε

 

That is actually fairly straightforward.

 

Try lim_(x->4) 1/2 (3x - 1) = 11/2 using  ε-δ.

 

To determine what δ should be given a particular ε, we start with:

 

| 1/2 (3x - 1) - 11/2 | < ε and work backwards to find δ in terms of ε.

 

Factor out 1/2:

 

1/2 | (3x-1) -11 | < ε

 

Multiply by 2:

 

| 3x - 12 | < 2ε

 

Factor out a 3:

 

3 | x - 4 | <  2ε

 

Divide by 3:

 

| x - 4 | <  2ε/3

 

There you have it. For any given ε, let δ = 2ε/3.

 

In other words I can get f(x) as close to 11/2 as I want, by letting x be within 2/3 of that distance.

 

The proof would go like this.

 

If  | x - 4 | < δ, then | 1/2 (3x-1) - 11/2 | < ε.

 

Let δ = 22ε/3/3.

 

 | x - 4 | < 2ε/3

 

| 3x - 12| < 2ε

 

1/2 |3x - 12| < ε

 

1/2 | 3x - 1 - 11 | < ε

 

| 1/2 (3x - 1) - 11/2 | < ε

 

Q.E.D.

 

Hopefully that gives you the idea.