Explanation and clarifications in epsilon-delta limit proofs

[bamajon1974]

TL;DR

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.

 

[anuttarasammyak]

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|&lt;|x-2|+4&lt;\delta+4
So
|x^2-4|&lt;\delta(\delta+4)&lt;\epsilon
is the required condition of $\delta$.

 

[PeroK]

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.

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.

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:
$$|x - 1| < 1 \ \Rightarrow \ |x| < 2 \ \Rightarrow \ |x^2 + 4x + 4| \le |x^2| + |4x| + 4 < 4 + 8 + 4 = 16$$And, 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!

 

[bamajon1974]

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

 

[PeroK]

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

It might be too late now.