Proving Limits Using Delta-Epsilon Method

[haushofer]

Hi,

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

<br /> \lim_{x \rightarrow 3} x^2 = 9<br />

So this means that I should prove that

<br /> |x-3| &lt; \delta \ \rightarrow \ \ |x^2 - 9| &lt; \epsilon(\delta)<br />

So I had the following idea:

<br /> |x^2 - 9| = |x-3||x+3|<br />

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

<br /> |x+3| = |x-3+6| &lt; |x-3| + 6<br />

So I get in total

<br /> |x-3| &lt; \delta \ \rightarrow \ |x^2 - 9| = |x-3||x+3| &lt; |x-3|(|x-3| + 6 ) &lt; \delta(\delta + 6)<br />

So choosing

<br /> \epsilon = \delta(\delta + 6)<br />

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?

 

[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|&lt;9|x-3|. This way you can choose \delta = \epsilon/9.

 

[haushofer]

I would get

<br /> \delta = \frac{-6 \pm \sqrt{36 + 4 \epsilon}}{2}<br />

If I take the positive sign I get

<br /> \delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}<br />

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

<br /> |x-3| &lt; \delta \ \ \ \ AND \ \ \ \ |x-3|&lt;3<br />

 

[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.

 

[HallsofIvy]

Hi,

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

<br /> \lim_{x \rightarrow 3} x^2 = 9<br />

So this means that I should prove that

<br /> |x-3| &lt; \delta \ \rightarrow \ \ |x^2 - 9| &lt; \epsilon(\delta)<br />

So I had the following idea:

<br /> |x^2 - 9| = |x-3||x+3|<br />

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

<br /> |x+3| = |x-3+6| &lt; |x-3| + 6<br />

So I get in total

<br /> |x-3| &lt; \delta \ \rightarrow \ |x^2 - 9| = |x-3||x+3| &lt; |x-3|(|x-3| + 6 ) &lt; \delta(\delta + 6)<br />

So choosing

<br /> \epsilon = \delta(\delta + 6)<br />

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

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?

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

 

[haushofer]

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

 

[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:

<br /> <br /> \delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}<br /> <br />

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

 

[Wizlem]

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?

Whether |x-3|&lt;3 is imposed or not does not change the fact that |x-3|&lt;\delta. These statements do not contradict each other, you still end up with |x-3|&lt;min(\delta,3)\leq\delta.

 

[haushofer]

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:

<br /> <br /> \delta = \frac{-6 + \sqrt{36 + 4 \epsilon}}{2}<br /> <br />

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

In my openingpost I had

<br /> <br /> \epsilon = \delta(\delta + 6)<br /> <br />

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.

 

[haushofer]

Whether |x-3|&lt;3 is imposed or not does not change the fact that |x-3|&lt;\delta. These statements do not contradict each other, you still end up with |x-3|&lt;min(\delta,3)\leq\delta.

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 ***.