# Proving the limit (epsilon delta) #3

1. Oct 12, 2009

### zeion

1. The problem statement, all variables and given/known data

Prove the following states directly using the formal e, d definition

$$\lim_{x\rightarrow 8} \sqrt{x + 1} = 3$$

2. Relevant equations

3. The attempt at a solution

If 0 < |x-8| < d
Then 0 < sqrt((x+1) - 3) < e

Let e be given
3 < sqrt(x+1) < e + 3
9 < x + 1 < e2 + 6e + 9
0 < x - 8 < e2 + 6e

This suggests we choose d = e2 + 6e

2. Oct 12, 2009

### Hurkyl

Staff Emeritus
I assume you want comments on your work? I have three, of varying degrees of importance.

1. The work you've done here is "working backwards" -- you've started with what you wanted to happen, and worked backwards to try and find something that would ensure it.

Now, what you want to do is to show that this choice of d really does do what you want it to do.

2. When you did this step:
did you remember to make sure everything was positive? A relevant law of arithmetic is
If 0 < a < b then a² < b²​

3. Your translation from limit to e-d has an error.

3. Oct 12, 2009

### zeion

3 < sqrt(x+1) < e + 3
9 < x + 1 < e2 + 6e + 9

Does this mean I need to show that sqrt(x+1) and e + 3 are positive?

Are you talking about the sqrt((x+1) - 3)? It should be sqrt(x+1) + 3?

4. Oct 12, 2009

### Hurkyl

Staff Emeritus
And 3. (But notice that if you can show 3 is positive, then it's very easy to show the other two are positive)

It's obvious enough in this case that you don't need to bother mentioning it in your work -- but I want to make sure that you know what you're doing when you apply steps like this.

"sqrt(x+1) - 3", actually.

5. Oct 12, 2009

### zeion

Yes I meant that sorry.

So now I need to show that
if 0 < x - 8 < e^2 + 6e
then 0 < sqrt((x+1) - 3) < e
is true?

6. Oct 12, 2009

### Hurkyl

Staff Emeritus
(did you mean "|x-8|"?)

If you are still interested in finishing the e-d problem you originally wrote down, then yes that is what you need to do.

Sometimes it's possible to argue that all of your scratch work is reversible -- but it's probably not worth trying to do so until you become very comfortable with doing these arithmetic proofs involving inequalities.