Delta-epsilon method

1. Sep 27, 2010

haushofer

Hi,

I had the following question of a student this day about proving the following limit:

$$\lim_{x \rightarrow 3} x^2 = 9$$

So this means that I should prove that

$$|x-3| < \delta \ \rightarrow \ \ |x^2 - 9| < \epsilon(\delta)$$

So I had the following idea:

$$|x^2 - 9| = |x-3||x+3|$$

The first term on the RHS is smaller than delta. For the second term I write

$$|x+3| = |x-3+6| < |x-3| + 6$$

So I get in total

$$|x-3| < \delta \ \rightarrow \ |x^2 - 9| = |x-3||x+3| < |x-3|(|x-3| + 6 ) < \delta(\delta + 6)$$

So choosing

$$\epsilon = \delta(\delta + 6)$$

should prove the statement, right? But if I see the book they're using (Apostol) I see that they "choose" |x-3|<C, try to make an inequality then for |x+3| etc. But for me that seems making things more difficult then they are. So my question is: is my answer described above right?

2. Sep 27, 2010

Wizlem

Hi,

I'm new to this forum so please excuse me if I mess up the latex. For the most part your method is sound. However, delta should be a function of epsilon since delta must exist for all epsilon. With the way you've done things getting delta as a function of epsilon is a bit messy. I would suggest limiting yourself to an interval around 3. Doing this with the interval (0,6) allows you to simply say |x+3| < 9 and you get $$|x^2 - 9| = |x-3||x+3|<9|x-3|$$. This way you can choose $$\delta = \epsilon/9$$.

3. Sep 28, 2010

haushofer

I would get

$$\delta = \frac{-6 \pm \sqrt{36 + 4 \epsilon}}{2}$$

If I take the positive sign I get

$$\delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}$$

For this particular delta it's true that it exists for all epsilon>0, right?

However, aren't you getting then an extra condition on |x-3|? So |x-3| is not only smaller than delta anymore in your proof; you also get |x-3|<3, right? So you get

$$|x-3| < \delta \ \ \ \ AND \ \ \ \ |x-3|<3$$

Last edited: Sep 28, 2010
4. Sep 28, 2010

haushofer

Maybe it's a matter of taste, but I find these extra conditions (limiting first to a certain interval) more messy than using the Cauchy-Schwarz inequality.

5. Sep 28, 2010

HallsofIvy

You are NOT allowed to choose $\epsilon$. You are given $\epsilon$ and want to choose $\delta$ to give that $\epsilon$.

No, it does not "prove the statement". You are going the wrong way.

6. Sep 28, 2010

haushofer

Why I'm going the wrong way? I've now found that
$$|x-3| < \delta \ \rightarrow \ \ |x^2 - 9| < \epsilon(\delta)$$
where the arrow is read as "implies", for the delta
$$\delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}$$

Last edited: Sep 28, 2010
7. Sep 28, 2010

Tac-Tics

You have written ε and a function of δ. You have it reversed. δ must be written as a function of ε.

"For every ε, there exists a δ such that...."

Because ε is in the outermost scope, δ may be written in terms of ε. However, since ε is a constant with respect to the rest of the expression, what you have above is incorrect.

But you're pretty close. In particular, if you haven't made a mistake in your algebra (I haven't checked), then you can easily turn:

$$\delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}$$

into an equation of the form "ε = ..."

8. Sep 28, 2010

Wizlem

Whether $$|x-3|<3$$ is imposed or not does not change the fact that $$|x-3|<\delta$$. These statements do not contradict eachother, you still end up with $$|x-3|<min(\delta,3)\leq\delta$$.

Last edited: Sep 28, 2010
9. Sep 29, 2010

haushofer

$$\epsilon = \delta(\delta + 6)$$

directly. I have to think about this more, but at this point I really don't see why I cannot use this epsilon to complete the proof.

10. Sep 29, 2010

haushofer

Ok, that's what I understand, but not why my approach in my openingspost is wrong. Sorry if I'm being a pain in the ***.

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