# Homework Help: Another epsilon-delta proof

1. Oct 1, 2007

### bjgawp

1. The problem statement, all variables and given/known data
Prove that $$\lim_{x\to\\-1}(3-2x^{2})=1$$ The problem isn't figuring out the problem but rather a question about the answer itself.[/

2. Relevant equations
None really.

3. The attempt at a solution
http://img248.imageshack.us/img248/913/prooffi5.jpg [Broken]

My question is that what's the difference if we assumed $$\delta$$ to be another number such as 5? We would wind up with $$\delta$$ = min{5, $$\epsilon$$/14}. How would the two answers differ? Is one answer more accurate or more "right" than the other?

Last edited by a moderator: May 3, 2017
2. Oct 1, 2007

### HallsofIvy

If $\delta$ is "assumed to be another number such as 5" then it wouldn't work would it? The whole point is to make certain that |2- 2x2|< 1 isn't it?

3. Oct 1, 2007

### Hurkyl

Staff Emeritus
Was that supposed to say $|2 - 2x^2| < \epsilon$?

I also assume you meant to say "If we had instead assumed $\delta \leq 5$..."? If you get an answer, then you get an answer. The answer is not unique.

For the record, that image has a minor logic error in it. And even if we ignore that, it's not a complete solution.

4. Oct 1, 2007

### bjgawp

Alright instead of $$\delta$$ = 5, what about $$\delta \eq$$ = 0.5? We would get $$\delta$$ = min{0.5, $$\epsilon$$/5}. Again, is this answer more "valid" than $$\delta$$ = min{1, $$\epsilon$$/6}

EDIT: Missed Hurkyl's post. Thanks again! Yeah, I didn't feel like going through the whole formality of the proof. Just wanted to get the point across. Thanks!

Last edited: Oct 1, 2007
5. Oct 1, 2007

### Hurkyl

Staff Emeritus
Like I said, if you get an answer, then you get an answer! This sort of technique is rather forgiving; it's rather common that choosing any number between 0 and $+\infty$ for your bound on $\delta$ will lead you to an answer.

6. Oct 1, 2007

### bjgawp

Just out of curiosity, what minor logic error did you refer to? Might help me solve a particular problem I've been having with these proofs.

Where I've said 2|x+1||x-1| < 2(3)|x+1| < $$\epsilon$$, why is it that we assume 2(3)|x+1| < $$\epsilon$$? Seems like we arbitrarily assumed it to make the solution work ... Thanks in advance guys!

Last edited: Oct 1, 2007
7. Oct 1, 2007

### Hurkyl

Staff Emeritus
Good call; that's exactly what I was referring to.

The work in this image -- the "working backwards" preliminary work -- is assuming the conclusion: $|2 - 2x^2| < \epsilon$. From our other assumption, we also know that $|2 - 2x^2| < 2(3)|x+1|$. However, we cannot conclude that $2(3)|x + 1| < \epsilon$. (At least, not immediately)

What we need here is, as you've realized, another assumption. (Though if we make the assumption so that the solution will work, it's not exactly arbitrary. )

We made the assumption that $\delta \leq 1$ so that we could control the size of $|x - 1|$. Now, since $2(3)|x+1|$ has appeared naturally, we make the additional assumption that $2(3)|x+1| < \epsilon$ to wedge it into the problem right where we want it.

By making these assumptions, we arrive at an answer that works, so we're happy! But if we were unable to arrive at an answer, we would have to try something else.

Remember that none of this work is the actual solution -- it's the process of working backwards from the answer to figure out what to use as a starting point. Once we have a starting point (e.g. "let $\delta = \min\{1, \epsilon/6\}$"), then we can work forwards to prove that this really is a solution.

Compare with solving algebra problems -- you manipulate the problem until you get potential solutions, and then you have to plug them back into the original equation to see if they really are solutions. The plugging-in part what you need to do to prove they are solutions -- but the manipulations are the process you used to figure out what the solutions had to be.

Last edited: Oct 1, 2007
8. Oct 1, 2007

### bjgawp

Thank you very much! Things are a bit more clearer than they were. However, just another question. I'm still not see how assuming 2(3)|x + 1| < $$\epsilon$$ leads us to find the "right" $$\delta$$. Just because $$\delta \leq$$ 1, where's our basis that even with the restriction we placed on |x - 1| that 2(3)|x+1| should be < $$\epsilon$$. Just curious where the connection is :) Thanks, once again, for taking your time!

9. Oct 1, 2007

### Hurkyl

Staff Emeritus
Your goal is to find a choice of $\delta$ that makes $|2 - 2x^2| < \epsilon$.

You have already shown that if $\delta \leq 1$, then $|2 - 2x^2| < 2(3)|x+1|$.

This gives you a lead -- if you can arrange things so that both $\delta \leq 1$ and $2(3)|x+1| < \epsilon$ are true, then you have accomplished your goal.

So, to continue working backward, you make the additional assumption that $2(3)|x+1| < \epsilon$, and see if that helps you deduce anything.

You aren't guaranteed that this will help you find the solution -- but this is something to try. And often, the simple thing works!

10. Oct 1, 2007

### bjgawp

Ooh! That clears it up for me. Do you know of an example where that assumption does NOT work? You don't have to go through the trouble of finding one as you've helped a lot already. Thank you very much :) Now .. onto making sense of proving the limit laws ...

11. Oct 1, 2007

### Hurkyl

Staff Emeritus
This very problem is a good example.

Suppose you didn't think to make the assumption that $\delta \leq 1$. Then, in the process of working backwards, you might do

$$|2 - 2x^2| = 2 |1-x| |1+x| < 2 \delta |1 - x|.$$

Then, the natural thing to do would be to assume that $2 \delta |1 - x| < \epsilon$ and see what happens. Solving for delta, you get:

$$\delta < \frac{\epsilon}{2 |1 - x|}$$

Unfortunately, since x can be any value at all, $|1 - x|$ can be arbitrarily large, and this forces us to choose $\delta \leq 0$, which we cannot do.

Incidentally, after meeting this failure, the next step to realize "Okay, the problem is that I had no control on the size of $|1 - x|$" -- it is this chain of thought that leads you to make an assumption like $\delta \leq 1$.