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.
$\text{Claim}: \lim_{x\to2}(\frac{1}{x^2}) = \frac{1}{4}\\$.
$\text{WTS}: \forall \epsilon>0, \ \exists \delta>0 \ \text{such that}, \ \forall x \in \mathbb{R}, \ 0<|x-2|<\delta \implies |\frac{1}{x^2} - \frac{1}{4}|<\epsilon.\\$
$\text{Scratch Work}:$
$\text{Manipulate the implication} \ 0<|x-2|<\delta \implies |\frac{1}{x^2}-\frac{1}{4}|<\epsilon \ \text{to find} \ \delta.$
$\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} \ ($$|4x^2|=4x^2$$ \ \text{since} \ $$4x^2$$ \ \text{is always positive}).$
$\text{By assumption, an upper bound of} \ |x-2| \ \text{is} \ \delta.$
$\text{Need to find an upper bound on} \ \frac{|x+2|}{4x^2} \ \text{by making} \ \frac{|x+2|}{4x^2} < \frac{C}{4 \cdot D} \ \text{for some numbers} \ C \ \text{and} \\ D. \ \text{Then any} \ \delta \leq (\frac{4 \cdot D}{C}) \epsilon \ \text{will bound}\ \frac{|x+2|}{4x^2} \ \text{above}.$
$\text{Choose} \ \delta \leq 1.$
$\text{Then, to find} \ C: \ |x-2|<1 \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies \\ 3 < x+2 < 5 \implies -5 < 3 < x+2 < 5 \implies |x+2| < 5.$
$\text{Similarly, to find} \ D: \ |x-2|<1 \implies -1 < x-2 < 1 \implies 1 < x < 3 \\ \implies 1>\frac{1}{x}>\frac{1}{3} \implies 1>\frac{1}{x^2}>\frac{1}{9} \implies \frac{1}{9}<\frac{1}{x^2}<1 \implies -1 < \frac{1}{9}<\frac{1}{x^2}<1 \implies |\frac{1}{x^2}|<1 \implies \frac{|1|}{|x^2|}<1 \implies \frac{1}{x^2}<1.$
$\text{Thus} \ \frac{C}{4 \cdot D} \ \leq \ \frac{5}{4 \cdot 1} = \frac{5}{4}.$
$\text{Then} \ \frac{|x-2||x+2|}{4x^2}< \delta \cdot (\frac{5}{4}) = \epsilon.$
And $\epsilon = (\frac{5}{4})\delta.$
$\text{So} \ \delta \leq 1 \ \text{and} \ \delta \leq \ (\frac{4}{5}) \epsilon \ \text{at the same time}.$
$\text{Choose} \ \delta=min[1,(\frac{4}{5}) \epsilon].$
$\text{Proof}$:
$\text{Let} \ \epsilon\ >0.$
$\text{Choose} \ \delta=min[1,(\frac{4}{5}) \epsilon].$
$\text{Let} \ x \in \mathbb{R}. \ \text{Assume} \ 0 < |x-2| < \delta. \ \text{This implies} \ |x-2|< (\frac{4}{5}) \epsilon \ \text{and} \ |x-2| < 1.$
$\text{Hence} \ |x-2|<1 \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies 3 < x+2 < 5 \implies \\ -5 < 3 < x+2 < 5 \implies |x+2| < 5$
$\text{and} \ |x-2|<1 \implies -1 < x-2 < 1 \implies 1 < x < 3 \implies 1>\frac{1}{x}>\frac{1}{3} \implies \\ 1>\frac{1}{x^2}>\frac{1}{9} \implies \frac{1}{9}<\frac{1}{x^2}<1 \implies -1 < \frac{1}{9}<\frac{1}{x^2}<1 \implies |\frac{1}{x^2}|<1 \implies \frac{|1|}{|x^2|}<1 \implies \frac{1}{x^2}<1.$
$\text{Furthermore} \ |\frac{1}{x^2}-\frac{1}{4}|= |\frac{4-x^2}{4x^2}|=\frac{|4-x^2|}{|4x^2|}=\frac{|x^2-4|}{4x^2}=\frac{|(x-2)(x+2)|}{4x^2}=\frac{|x-2||x+2|}{4x^2}.$
$\text{Then} \ \frac{|x-2||x+2|}{4x^2}< \delta \cdot (\frac{5}{4}) \leq (\frac{4}{5}) \epsilon \cdot (\frac{5}{4}) = \epsilon.$
$\text{Thus} \ |\frac{1}{x^2}-\frac{1}{4}|<\epsilon. \ _\blacksquare$
$\text{Questions}:$
- Is the scratch work and proof correct?
- Have I used equal and inequality signs for delta correctly in the scratch work and proof?
- Am I using the bounding above terminology found in the scratch work correctly?
- (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 $|x+2|=C$ and $x^2=D$ terms individually, or am I supposed to bound the $\frac{|x+2|}{x^2}$ as a single quotient?
- (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!
