Proving the limit (epsilon delta) #3

[zeion]

Homework Statement

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

<br /> \lim_{x\rightarrow 8} \sqrt{x + 1} = 3<br />

Homework Equations

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

 

[Hurkyl]

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:

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

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.

 

[zeion]

did you remember to make sure everything was positive? A relevant law of arithmetic is

If 0 < a < b then a² < b²​

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?

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

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

 

[Hurkyl]

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?

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.

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

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

 

[zeion]

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

Yes I meant that sorry.

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

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

 

[Hurkyl]

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

(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.