Rational epsilon-delta limit proof questions

[bamajon1974]

Homework Statement

This post is not a request to "do my homework".

Relevant Equations

See post above.

Summary:: Good afternoon. I have more questions about the details of epsilon-delta proofs. Below is a simple, rational limit proof example with questions at the end. The scratch work and proof are a bit pedantic but I don't follow proofs very well which omit a lot of details, including scratch work and thinking processes.

Good afternoon. I have more questions about the details of epsilon-delta proofs. Below is a simple, rational limit proof example with questions at the end. The scratch work and proof are a bit pedantic but I don't follow proofs very well which omit a lot of details, including scratch work and thinking processes.

$\text{Claim}: \lim_{x\to2}(\frac{1}{x^2}) = \frac{1}{4}\\.$

$\text{WTS}: \forall \epsilon>0, \exists \delta>0 \ \text{such that}, \forall x \in \mathbb{R}, 0<|x-2|<\delta \implies |\frac{1}{x^2} - \frac{1}{4}|<\epsilon\\.$

$\text{Scratch Work:}$

- $\text{Manipulate implication} \ 0<|x-2|<\delta \implies |\frac{1}{x^2}-\frac{1}{4}|<\epsilon \ \text{to find} \ \delta.$

- $\text{So} \ |\frac{1}{x^2}-\frac{1}{4}|= |\frac{4-x^2}{4x^2}|=\frac{|4-x^2|}{|4x^2|}=\frac{|x^2-4|}{4x^2}=\frac{|(x-2)(x+2)|}{4x^2}=\frac{|x-2||x+2|}{4x^2}.$

- $\text{By assumption, an upper bound of} \ |x-2| \ \text{is} \ \delta.$

- $\text{Need to find an upper bound on} \ \frac{|x+2|}{4x^2} \ \text{term by making} \ \frac{|x+2|}{4x^2} < \frac{C}{4 \cdot D} \ \text{for some numbers} \ C \ \text{and} \$ $D. \ \text{Then any} \ \delta \leq (\frac{4 \cdot D}{C}) \epsilon \ \text{will bound} \ \frac{|x+2|}{4x^2} \text{above}.$

- $\text{Choose} \ \delta \leq 1.$

- $\text{Then, to find} \ C: \ |x-2|<\delta \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies 3 < x+2 < 5$ $\implies -5 < 3 < x+2 < 5 \implies |x+2| < 5.$

- $\text{Similarly, to find} \ D: \ |x-2|<\delta \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies 1>\frac{1}{x}>\frac{1}{3}$ $\implies 1>\frac{1}{x^2}>\frac{1}{9} \implies \frac{1}{9}<\frac{1}{x^2}<1 \implies -1 < \frac{1}{9}<\frac{1}{x^2}<1 \implies \frac{1}{|x^2|}<1 \implies \frac{1}{x^2}<1.$

- $\text{Thus} \ \frac{C}{4 \cdot D} \ \leq \ \frac{5}{4 \cdot 1} = \frac{5}{4}.$

- $\text{Then} \ \frac{|x-2||x+2|}{4x^2}< \delta \cdot (\frac{5}{4}) = \epsilon.$

- $\text{So} \ \delta \leq 1 \ \text{and} \ \delta \leq (\frac{4}{5}) \epsilon \ \text{at the same time}.$

- $\text{Choose} \ \delta=min[1,(\frac{4}{5}) \epsilon].$

$\text{Proof:}$

- $\text{Let} \ \epsilon\ >0.$

- $\text{Choose} \ \delta=min[1,(\frac{4}{5}) \epsilon].$

- $\text{Let} \ x \in \mathbb{R}. \ \text{Assume} \ 0 < |x-2| < \delta. \ \text{This implies} \ |x-2|< (\frac{4}{5}) \epsilon \ \text{and} \ |x-2| < 1.$

- $\text{Hence} \ |x-2|<\delta \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies 3 < x+2 < 5 \implies$ $-5 < 3 < x+2 < 5 \implies \\ |x+2| < 5$

- $\text{and} \ |x-2|<\delta \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies 1>\frac{1}{x}>\frac{1}{3} \implies$ $1>\frac{1}{x^2}>\frac{1}{9} \implies \frac{1}{9}<\frac{1}{x^2}<1 \implies -1 < \frac{1}{9}<\frac{1}{x^2}<1 \implies \frac{1}{|x^2|}<1 \implies \frac{1}{x^2}<1.$

- $\text{Then} \ \frac{|x-2||x+2|}{4x^2}< (\frac{4}{5}) \epsilon \cdot (\frac{5}{4}) = \epsilon.$

- $\text{Thus} \ |\frac{1}{x^2}-\frac{1}{4}|<\epsilon. \ _\blacksquare$

Questions:

1. Is the scratch work and proof correct?
2. Have I used equal and inequality signs for delta correctly in the scratch work and proof?
3. Am I using the bounding above terminology found in the scratch work correctly?
4. (Most important question). Bullet #4 of the scratch work. When determining an upper bound for $\frac{|x+2|}{x^2}$ term, do I bound the $|x+2|$ and $x^2$ terms separately or $\frac{|x+2|}{x^2}$ together as a single quotient?
5. (I have forgotten quite a bit on manipulating absolute value inequalities over the years) Regardless of the answer to Q4, how would I manipulate the $\frac{|x+2|}{x^2}$ collectively (and not split the numerator and denominator) to arrive at an $|x-2|$ term? With or without using the triangle inequality?

Thank you!

 

[FactChecker]

Here is why you should be skeptical of your answer. When $\epsilon$ is very small, $\delta$ must become very small, much smaller than min(1, 5/4) = 1.
You should start with an arbitrarily small $\epsilon$ and find an equation for $\delta$ that depends on $\epsilon$.

 

[bamajon1974]

Here is why you should be skeptical of your answer. When $\epsilon$ is very small, $\delta$ must become very small, much smaller than min(1, 5/4) = 1.
You should start with an arbitrarily small $\epsilon$ and find an equation for $\delta$ that depends on $\epsilon$.

Woops. I found several errors. I was able to correct.

 

[FactChecker]

That is much better. It is the type of equation for $\delta$ that one should expect.
The part that you did as scratch work should be part of the proof. You can either work it into the main logic flow of the proof or state it as a lemma and prove the lemma (saying that if $|x-2|\lt \min{(1,5\epsilon/4)}$ then $|1/x^2 -1/4| \lt \epsilon$.) I think it should be worked into the main proof.

 

[bamajon1974]

That is much better. It is the type of equation for $\delta$ that one should expect.
The part that you did as scratch work should be part of the proof. You can either work it into the main logic flow of the proof or state it as a lemma and prove the lemma (saying that if $|x-2|\lt \min{(1,5\epsilon/4)}$ then $|1/x^2 -1/4| \lt \epsilon$.) I think it should be worked into the main proof.

Thank you.

A few follow up questions...The part that I did as the scratch work that you suggest including in the proof, do you mean the line where I show $|\frac{1}{x^2} - \frac{1}{4}| = \frac{|x-2||x+2|}{4x^2}$ ? I would think the best place to show this work would be between the 5th and 6th bullet points of the proof, right?

In your 2nd reply, you mention, $|x-2|\lt \min{(1,5\epsilon/4)}$, shouldn't it be $|x-2|\lt \min{(1,4\epsilon/5)}$ instead?

Finally, I still have the question about how would I transform $\frac{|x+2|}{4x^2}<1$ as a whole into $|x-2|<\delta$ rather than individually manipulate the numerator and denominator separately. Not exactly sure how to do this.

Thanks!

 

[PeroK]

In your 2nd reply, you mention, $|x-2|\lt \min{(1,5\epsilon/4)}$, shouldn't it be $|x-2|\lt \min{(1,4\epsilon/5)}$ instead?

Yes.

Finally, I still have the question about how would I transform $\frac{|x+2|}{4x^2}<1$ as a whole into $|x-2|<\delta$ rather than individually manipulate the numerator and denominator separately. Not exactly sure how to do this.

I'm not sure why you would want to do this. You have $|x - 2| < 1 \ \Rightarrow \ \frac{|x + 2|}{4x^2} < \frac 5 4$, which is good enough.

 

[anuttarasammyak]

To check your way please find below another approach.-\epsilon&lt;\ \frac{1}{x^2}- \frac{1}{4}\ &lt;+\epsilon
\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}&lt;x&lt;\sqrt{\frac{1}{\frac{1}{4}-\epsilon}}
-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}&gt;x&gt;-\sqrt{\frac{1}{\frac{1}{4}-\epsilon}}
We are focusing on positive x. For both the larger and the smaller inequalities hold,
|x-2|&lt;\min (2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}, \sqrt{\frac{1}{\frac{1}{4}-\epsilon}}-2)=2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}
So we shall introduce $\delta$ such as
|x-2|&lt;\delta&lt;2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}

[EDIT]$\epsilon<\frac{1}{4}$ so that number within the square root is positive.

 

[bamajon1974]

To check your way please find below another approach.-\epsilon&lt;\ \frac{1}{x^2}- \frac{1}{4}\ &lt;+\epsilon
\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}&lt;x&lt;\sqrt{\frac{1}{\frac{1}{4}-\epsilon}}
-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}&gt;x&gt;-\sqrt{\frac{1}{\frac{1}{4}-\epsilon}}
We are focusing on positive x. For both the larger and the smaller inequalities hold,
|x-2|&lt;\min (2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}, \sqrt{\frac{1}{\frac{1}{4}-\epsilon}}-2)=2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}
So we shall introduce $\delta$ such as
|x-2|&lt;\delta&lt;2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}

This is an interesting way of finding delta in terms of epsilon, seems more general. Does this technique find the optimal delta range that works, as opposed to say, delta=min{1, 5*epsilon/4) which is just a delta range that works but not necessarily the optimal range?

Also, in the second to last step, why do you keep juts the 2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}} term and not both?

How would I check my approach in the proof above? I am unfamiliar with your technique.

Thanks!

 

[anuttarasammyak]

Also, in the second to last step, why do you keep juts the 2−114+ϵ term and not both?

How would I check my approach in the proof above? I am unfamiliar with your technique.

Say |x-2| < a < b, |x-2| < a is enough for us to introduce $\delta$ so that |x-2| <$\delta$ < a.
If the results of two approaches do not contradict, so if your upper estimate of $\delta$ is $\frac{5\epsilon}{4}$,
2-\sqrt{\frac{1}{\frac{1}{4}+\epsilon}}=\frac{8\epsilon}{\sqrt{1+4\epsilon}(\sqrt{1+4\epsilon}+1)}&gt;\frac{8\epsilon}{\sqrt{2}(\sqrt{2}+1)}&gt;2.9\epsilon&gt;\frac{5\epsilon}{4}
holds for $\epsilon< \frac{1}{4}$. We are encouraged to be on a right track.

 

[PeroK]

Does this technique find the optimal delta range that works ...

Finding the "optimal" delta is a waste of time and effort. You may be able to do it in the quadratic case, but what about a cubic case? Or, higher polynomials?

Instead, you should be trying to learn techniques that give you a suitable delta with the least effort. There are no prizes for getting an "optimal" delta.