Question

Limit Question:
Okay, so using the DEFINITION of a limit prove x^2-2x-1=-2.

Here's the issue I have with it. I understand the formula completely, I get how to do it, but in my advanced math class I have to create a graph of limits using Maple and I can't figure out how to do it when you break it up and you get |x-1||x-1|...it just doesn't make sense to me. Any help? Also, if you don't understand I'll try to make it clearer lol

Answer 1

I'm a bit confused here... x^2-2x-1=-2 is only valid for x = 1.

But x^2-2x-1 can also be equal to any value greater/equal to -2.

How do you "prove" that x^2-2x-1=-2 when it's not even true unless x=1? Maybe I need to look at the definition of a limit again...

Answer 2

Here lemme explain it a little bit better:
lim(x->1): x^2 - 2x - 1 = -2
So, when you prove it it turns into this:
|(x^2 - 2x - 1) + 2|=|x^2 - 2x + 1|=|x-1||x-1|

What I'm confused about it the next step in terms of a formal definition. Usually when you have a quadratic equation you have (-) and a (+) number so you have to bound the higher value but the solutions in this case are the same! So I have no idea how to continue.

Does this make more sense?

Answer 3

Normally, I'd try to solve it with two cases. One case being when some \[\epsilon\] is less than my biggest bound (I use 1/2) and then another case being when some epsilon is greater than that bound. But when it's the same number, it's confusing.

Answer 4

i will help u in a bit, i just have a phone call, ur problem is sorted

Answer 5

awesome! thanks!

Answer 6

doesn't |x-1||x-1| < e just become
|x-1|^2 < e, so
|x-1| < sqrt(e), then
1 - sqrt(e) < x < 1 + sqrt(e) ?

Answer 7

In all of my examples I would never sqrt(e)....I'm very confused with why you did that. I don't even understand how you'd continue, BUT I will try it first to see if I get anything helpful before I criticize :)

Answer 8

yeah, I know it seems weird, but since the error term is always greater than 0 taking the square root is mathematically sound at least.

Answer 9

I tried it but I can't see how it works with the formula.
|x-1|^2 < delta
|x-1| < sqrt(delta)
So
|x-1|*sqrt(delta)

Idk that just doesn't seem right at all...

Answer 10

Well, I don't think we would use |x-1|^2 < delta. For the proof of a limit don't we use:
|x - limit| < delta?

So since our limit is 1, we have |x - 1| < delta.

But we can just choose delta = sqrt(e) since we know that |x-1| < sqrt(e).

That proves that there exists a delta that will satisfy the error formula.

Answer 11

That is, for any given error bound, just choose the square root of the error for delta and you will stay within the bounds.

Answer 12

But we don't know |x-1| < sqrt(e) yet....because you can't just say that delta = sqrt(e)...can you?
Okay, this squaring this is confusing me so much now lol

Answer 13

OK, we know that |x - 1| < sqrt(e) because we prove that as follows:

By definition,

|Function - Limit Value| < e
|x^2 -2x - 1+ 2| < e
|x^2 - 2x + 1| < e
|(x-1)(x-1)|<e
|(x-1)^2|<e
|x-1|^2<e
|x-1|<sqrt(e)

This proves that for our limit to work it must be such that |x-1| < sqrt(e).

Answer 14

Now, we can ensure that this is the case by turning to the second part of the definition:

|x - Limit| < delta
|x - 1| < delta

We need only find ANY delta that will make the equation work. So we know that if |x-1| < sqrt(e) it will work. So just pick a delta that works, namely, delta = sqrt(e).

To prove that a limit exists, you only need to show that there exists SOME, ANY value of delta that satisfies the error equation.

Answer 15

Okay, well I'm exhausted. But I will look at this again tomorrow. Thanks for your help!

Answer 16

no problem, good luck!

Answer 17

x^2-2x-1
=X^2 -2x +1 -2
=(x-1)^2 -2
here (x-1)^2 is always positive
hence the least value of (x-1)^2 is 0
hence the least value of the expression x^2-2x-1 is -2

Answer 18

The work that @jtvatsim has shown is absolutely correct. However, I must point out that the definition of the limit does not operate in that way. Your proof would be sufficient for ordinary professors, but possibly not all. The reason is that by definition:
lim (x =>x1) f(x) = L if and only if
for every epsilon > 0, there exists a delta > 0 such that
|f(x) - L| < epsilon whenever |x-x1|<delta.

The keyword here is whenever. A whenever B is not A implies B but rather B implies A. Your work is merely a way of estimating delta in terms of epsilon. the complete, foolproof proof requires that you show
|x-x1| < delta => |f(x)-L| < epsilon
when for delta as some function of epsilon.

Answer 19

you want an \(\epsilon, \delta\) proof right?

Answer 20

and you have reached the step where you want a \(\delta\) so that
\[|x^2-2x+1|<\epsilon\]

Answer 21

and have seen that this is the same as wanting
\[|x-1||x-1|=|x-1|^2<\epsilon\] now you pick the \(\delta\) in terms of \(\epsilon\) and work the algebra backwards is all you need to do. that is, pick \(\delta=\sqrt{\epsilon}\)

Answer 22

too much work
if \(\delta=\sqrt{\epsilon}\) then given \(|x-1|<\sqrt{\epsilon}\) you have
\[|x-1||x-1|<\sqrt{\epsilon}^2=\epsilon\] whic is all you need

Answer 23

all the work is finding the appropriate \(\delta\) in terms of \(\epsilon\)
the "proof" is just working the algebra backwards

Answer 24

This is my "finished" proof. I think I finally got it. 'Think' being the key word ;)

Answer 25

\[Choose \delta = \sqrt{\epsilon}\]
By the definition of the limit,
\[|x-1| < \delta = \sqrt{\epsilon}\]
which implies
\[|x-1|^{2} < \epsilon\]
which implies
\[|(x-1)^{2}| = |x ^{2}-2x+1| = |(x ^{2}-2x -1) - (-2)| = |f(x) - L|< \epsilon\].
QED.
There is no need to consider cases.

Answer 26

The only reason I'm doing cases is because that's what my professor told me to do. I don't understand why I'd 'have' to but I'm doing it to appease her.

Answer 27

Do you know why she is doing so? That's actually pretty odd. There is no reason why 1/2 should be important. epsilon = 1/4 has virtually no significance in this limit proof.

Answer 28

In any case, your case 2 proof isn't quite right as you need to replace one of the equality signs with inequality.

The formal proof of this limit doesn't require cases. Many problems do, but this isn't one of them.