- #1

bamajon1974

- 21

- 5

- 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!

Last edited: