# Proving Limits Using Delta-Epsilon Method

• haushofer
In summary, the conversation is about proving the limit of x^2 as x approaches 3. The approach involves using the epsilon-delta definition and using the Cauchy-Schwarz inequality. However, there is a disagreement about whether delta should be chosen as a function of epsilon or the other way around. There is also a discussion about whether the condition |x-3|<3 is necessary in the proof.
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?

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 $$\delta = \epsilon/9$$.

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:
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.

haushofer said:
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

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

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:
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 = ..."

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

Last edited:
Tac-Tics said:
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 = ..."

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

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.

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

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 ***.

## 1. What is the Delta-Epsilon Method?

The Delta-Epsilon Method is a rigorous mathematical technique used to prove the existence of limits in calculus. It involves using the concepts of delta (a small change in the input) and epsilon (a small error or tolerance) to show that for any epsilon, there exists a corresponding delta such that the output of a function will be within epsilon of the limit as the input approaches a given value.

## 2. Why is the Delta-Epsilon Method important?

The Delta-Epsilon Method is important because it provides a solid mathematical foundation for the concept of limits. It allows us to prove the existence of limits in a precise and rigorous way, providing a solid basis for further calculations and applications in calculus and other areas of mathematics.

## 3. How is the Delta-Epsilon Method used to prove limits?

The Delta-Epsilon Method involves setting up an inequality that relates the input value, delta, and epsilon, and then manipulating the inequality to find an expression for delta in terms of epsilon. This expression can then be used to show that for any given epsilon, there exists a corresponding delta that satisfies the requirements for the limit to exist.

## 4. Are there any limitations to the Delta-Epsilon Method?

Like any mathematical technique, the Delta-Epsilon Method has its limitations. It can only be used to prove the existence of limits, not to calculate them. It also requires careful attention to detail and can be quite time-consuming, making it less practical for certain applications.

## 5. How can I improve my understanding of the Delta-Epsilon Method?

To improve your understanding of the Delta-Epsilon Method, it is important to practice solving a variety of problems using this technique. You can also seek out additional resources, such as textbooks or online tutorials, and work through examples and exercises. Additionally, discussing the method with your peers or a tutor can help clarify any confusing concepts.

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