# Explanation and clarifications in epsilon-delta limit proofs

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• bamajon1974
In summary: Set$$\delta = \min\left(1,\frac{\epsilon}{11}\right)$$Then$$0 < |x - 5| < \delta \ \Rightarrow \ |x + 5| < 11$$So$$0 < |x - 5| < \delta \ \Rightarrow \ 0 < 3|x-5||x+5| < 3\delta \cdot 11 = 33\delta$$Now$$\delta < 1 \ \Rightarrow \ 33\delta < 33$$Therefore$$0 < |x - 5| < \delta \ \Rightarrow \ 0 < 3|x-5||x+5| < bamajon1974 TL;DR Summary I have some questions about the details of epsilon-delta proofs. The questions are below the example and involve clarification and explanations of steps and details in the scratch work. Good afternoon. I have some questions about the details of epsilon-delta proofs. Below is a simple, non-linear limit proof example which will serve as an example of the questions I have. The questions are below the example and involve clarification and explanations of steps and details in the scratch work. Prove$$\lim_{x \to 2} x^2 =4$$Want to show$$\forall \epsilon > 0, \exists \delta > 0 | \forall x \in \mathbb{R}, 0 < |x-2| < \delta \implies |x^2-4| < \epsilon$$Scratch Work (to find ##\delta##) - Manipulate implication ##0 < |x-2| < \delta \implies |x^2-4| < \epsilon## to find ##\delta##. - Then ##|x^2-4| = |(x+2)(x-2)| = |x+2||x-2| < |x+2|\cdot\delta##. - What to do with ##|x+2|## term? ##\delta## cannot depend on ##x##, only ##\epsilon##. - Establish upper bound on ##|x+2|## term by making ##|x+2| < C## for some number ##C##, then any ##\delta \leq \frac{\epsilon}{C}## will work. - Choose ##\delta \leq 1##. Then ##|x-2|<1 \implies -1<x-2<1 \implies 1<x<3 \implies 3<x+2<5 \implies## ##-5<3<x+2<5 \implies |x-2|<5 ##. - Alternatively (using triangle inequality theorem), choose ##\delta \leq 1##. Then ##|x-2| < 1##. Now ##|x+2| = |x-2+4| \leq |x-2| + |4| = |x-2| + 4 < 1+4 = 5##. - Then ##|x+2|\cdot\delta = 5\cdot\delta##. - So ##\delta \leq 1## and ##\delta \leq \frac{\epsilon}{5}## at the same time. Take ##\delta = \min[1,\frac{\epsilon}{5}]##. Actual Proof Claim:$$\forall \epsilon > 0, \exists \delta > 0 | \forall x \in \mathbb{R}, 0 < |x-2| < \delta \implies |x^2-4| < \epsilon$$Proof: - Let ##\epsilon > 0##. - Take ##\delta = \min[1,\frac{\epsilon}{5}]##. - Let ##x \in \mathbb{R}##. Assume ##0<|x-2|<\delta##. This implies ##|x-2|<\frac{\epsilon}{5}## and ##|x-2|<1##. - Hence ##|x-2|<1 \implies -1<x-2<1 \implies 1<x<3 \implies 3<x+2<5 \implies -5<3<x+2<5## ##\implies |x-2|<5 ##. - Then ##|x^2-4| = |(x+2)(x-2)| = |x+2||x-2| < (\frac{\epsilon}{5})\cdot5 = \epsilon##. - Thus ##|x^2-4| < \epsilon. \blacksquare## Questions 1. Is my scratch work and proof correct? 2. Last line of scratch work. When ##\delta## is found should it be equal to or less than or equal to some values? ##\delta = \min[1,\frac{\epsilon}{5}]## *OR* ##\delta \leq \min[1,\frac{\epsilon}{5}]##? 3. Scratch work. Is the phrase "getting control" of ##|x+2|## term the same as establishing an upper bound? I hear the "getting control" phrase frequently and want to confirm. 4. Is there any geometric interpretation to accompany the algebraic manipulations for the process of establishing an upper bound of ##|x+2|## term? 5. Similarly to Q4. In the graph of ##y=x^2##, for some ##\delta## around ##x=2##, the distance between ##2## and ##2-\delta## is not the same as the distance between ##2## and ##2+\delta##. So, from the scratch work, when ##\delta = \min[1,\frac{\epsilon}{5}]## and the smaller of the two values is chosen, can this geometrically be interpreted as picking the smaller ##\delta## band distance previously mentioned? Or are the two concepts unrelated? 6. Can I get some clarification establishing upper bounds in the scratch work? Is an upper bound established for the entire function, ##y=x^2## itself or is the upper bound found on just the ##|x+2|## term (since ##|x-2|## is bounded by ##\delta##)? I am pretty sure the latter but want to confirm. Also, I understand the algebra of turning ##|x-2|<1## into ##|x+2|<5##. But how does one justify using the ##|x-2|## term to come up with an upper bound for ##|x+2|## term? 7. Please refer to the webpage Milefoot which demonstrates epsilon-delta proofs for non-linear functions. The author(s) use a seemingly different way to find delta. How does the method for finding delta in the scratch work above differ from that in the webpage? Or why is the author of the website calculating delta in that way? Just a hunch, but in Q5, would a geometric interpretation of the unequal delta bands in a non-linear function apply to the way delta is calculated and chosen from the website instead? Thank you for your help. Last edited: PeroK bamajon1974 said: Summary:: I have some questions about the details of epsilon-delta proofs. The questions are below the example and involve clarification and explanations of steps and details in the scratch work. - What to do with |x+2| term? δ cannot depend on x, only ϵ. $$|x+2|=|x-2+4|<|x-2|+4<\delta+4$$ So $$|x^2-4|<\delta(\delta+4)<\epsilon$$ is the required condition of ##\delta##. bamajon1974 said: Actual Proof Claim:$$\forall \epsilon > 0, \exists \delta > 0 | \forall x \in \mathbb{R}, 0 < |x-2| < \delta \implies |x^2-4| < \epsilon$$Proof: - Let ##\epsilon > 0##. - Take ##\delta = \min[1,\frac{\epsilon}{5}]##. - Let ##x \in \mathbb{R}##. Assume ##0<|x-2|<\delta##. This implies ##|x-2|<\frac{\epsilon}{5}## and ##|x-2|<1##. - Hence ##|x-2|<1 \implies -1<x-2<1 \implies 1<x<3 \implies 3<x+2<5 \implies -5<3<x+2<5## ##\implies |x-2|<5 ##. - Then ##|x^2-4| = |(x+2)(x-2)| = |x+2||x-2| < (\frac{\epsilon}{5})\cdot5 = \epsilon##. - Thus ##|x^2-4| < \epsilon. \blacksquare## This is fine, except several times you wrote ##|x - 2| < 5## instead of ##|x+2| < 5##. Also:$$|x - 2| < 1 \ \Rightarrow \ |x+2| = |x - 2+ 4| \le |x -2| + 4 < 5$$follows from the triangle inequality. You have definitely got the right idea about these proofs. bamajon1974 said: Questions 1. Is my scratch work and proof correct? 2. Last line of scratch work. When ##\delta## is found should it be equal to or less than or equal to some values? ##\delta = \min[1,\frac{\epsilon}{5}]## *OR* ##\delta \leq \min[1,\frac{\epsilon}{5}]##? It doesn't really matter, but technically you want a specific delta, so ##=## is better. bamajon1974 said: 7. Please refer to the webpage Milefoot which demonstrates epsilon-delta proofs for non-linear functions. The author(s) use a seemingly different way to find delta. How does the method for finding delta in the scratch work above differ from that in the webpage? Or why is the author of the website calculating delta in that way? Just a hunch, but in Q5, would a geometric interpretation of the unequal delta bands in a non-linear function apply to the way delta is calculated and chosen from the website instead? I don't like the way the author has done that, as there is no obligation to find precise deltas by solving quadratic equations. There are easier ways to find a good-enough delta. I'll show you what I mean. We need to show that$$\lim_{x \to 5}(3x^2 -1) = 74$$Proof: First, note that$$|(3x^2 -1) - 74| = 3|x^2 - 25| = 3|x-5||x + 5|$$Now$$0 < |x - 5| < 1 \ \Rightarrow \ |x + 5| < 11$$Let ##\epsilon < 0##, and take ##\delta = min(1,\frac{\epsilon}{33})##, then:$$0 < |x -5| < \delta \ \Rightarrow \ |(3x^2 -1) - 74| = 3|x-5||x + 5|< 33|x-5| < \epsilon$$Note that you cannot use the technique in that webpage for cubic or higher polyomials. For example, to show that:$$\lim_{x \to 1}(x^3 + 3x^2 -1) = 3$$we note that:$$|x^3 + 3x^2 -4| = |x -1||x^2 + 4x + 4|$$Then we need an upper bound for that quadratic when ##|x - 1| < 1##, say. Again, using the triangle inequality:$$|x - 1| < 1 \ \Rightarrow \ |x| < 2 \ \Rightarrow \ |x^2 + 4x + 4| \le |x^2| + |4x| + 4 < 4 + 8 + 4 = 16And, we see that ##\delta = min(1, \frac{\epsilon}{16})## should do the trick.

In fact, you can even see the beginnings of a proof that any polynonial function has the expected limit by using the remainder theorem and the boundedness of a polynomial on a finite interval. And prove the whole lot in one fell swoop!

I did find some mistakes. How do I correct the mistakes in the original post?

bamajon1974 said:
I did find some mistakes. How do I correct the mistakes in the original post?
It might be too late now.

## 1. What is an epsilon-delta limit proof?

An epsilon-delta limit proof is a method used in mathematics to formally prove the limit of a function as it approaches a certain value. It involves choosing a small value (epsilon) and showing that for any input within a certain distance (delta) from the limit, the output of the function will be within epsilon of the limit.

## 2. Why are epsilon-delta limit proofs important?

Epsilon-delta limit proofs are important because they provide a rigorous and precise way to prove the limit of a function. They are also used in many areas of mathematics, such as calculus and analysis, and are essential for understanding and solving more complex problems.

## 3. What is the role of epsilon and delta in these proofs?

Epsilon and delta are both variables used in epsilon-delta limit proofs. Epsilon represents a small value that the output of the function must be within of the limit. Delta represents a small distance from the limit, and the goal of the proof is to show that for any input within delta of the limit, the output will be within epsilon of the limit.

## 4. How do you choose the values for epsilon and delta in a limit proof?

The values for epsilon and delta are chosen based on the specific function and limit being evaluated. Typically, epsilon is chosen first and then delta is determined based on the chosen value of epsilon. The goal is to find a delta that works for all inputs within a certain distance from the limit.

## 5. What are some common mistakes to avoid in epsilon-delta limit proofs?

Some common mistakes to avoid in epsilon-delta limit proofs include using the same value for epsilon and delta, assuming that delta can be any value, and not considering all possible inputs within delta of the limit. It is important to carefully choose epsilon and delta and to consider all cases in the proof to ensure its validity.

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