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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!

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  • $\begingroup$ Corrected the errors. $\endgroup$ Commented Mar 5, 2022 at 3:37
  • $\begingroup$ We are only concerned with values of $x$ that are close to $2$, so we may restrict our attention to $\delta \le 1,$ that is, $|x-2|\le1. $ For all such values of $x,$ the numerator in $\frac {|x+2|}{4x^2}$ is $\le 3$ and the denominator is $\ge 4.$ So $|x-2|\le 1\implies |1/4-1/x^2|=|x-2|\cdot \frac {|x+2|}{4x^2}\le |x-2|\cdot\frac 3 4.$ $\endgroup$ Commented Mar 8, 2022 at 2:11

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I will comment on the proof. You should include the important intermediate computations in your proof. Here you should include $$\frac{1}{x^2} - \frac{1}{4} = \frac{4 - x^2}{4x^2} = \frac{(2 + x)(2 - x)}{4x^2}.$$ I would write this before any $\epsilon$-$\delta$ stuff. The moment you write this, your epsilon and delta stuff will be motivated and easily understood by the reader. The rest looks okay.

It is more natural to show that $\lim_{h \to 0}\frac{1}{(2 + h)^2} = \frac{1}{4}$. This has the advantage that the thing which you are allowed to make small, $h$, is put in a variable for you without you having to force it out.

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  • $\begingroup$ Thank you for the reply. You mean that add the line: $$\frac{1}{x^2} - \frac{1}{4} = \frac{4 - x^2}{4x^2} = \frac{(2 + x)(2 - x)}{4x^2} $$ before the Let $\epsilon>0$ line? Wouldn't it fit better between lines 5 and 6 of the proof? $\endgroup$ Commented Mar 7, 2022 at 16:40
  • $\begingroup$ Also, do you have an example proof demonstrating $\lim_{h \to 0} \frac{1}{(x+h)^2} = L$ ? I understand the reasoning, but am unfamiliar with a proof based on this method. $\endgroup$ Commented Mar 7, 2022 at 16:43
  • $\begingroup$ @bamajon1974 The statement "$\lim_{x \to 2}\frac{1}{x^2} = L$" is identical to the statement "$\lim_{h \to 0}\frac{1}{(2 + h)^2} = L$". $\endgroup$
    – Mason
    Commented Mar 7, 2022 at 19:59
  • $\begingroup$ @bamajon1974 Yes it could go between lines 5 and 6. But I like to put it first because it motivates the choice of $\delta$. It is the most important part of the proof. $\endgroup$
    – Mason
    Commented Mar 7, 2022 at 20:04
  • $\begingroup$ Done and done. I added the line and a little more justification to the proof. $\endgroup$ Commented Mar 7, 2022 at 23:25

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