# Rational Epsilon-Delta Limit Proof Questions

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!

• Corrected the errors. Commented Mar 5, 2022 at 3:37
• 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 $|x-2|\le 1\implies |1/4-1/x^2|=|x-2|\cdot \frac {|x+2|}{4x^2}\le |x-2|\cdot\frac 3 4.$ Commented Mar 8, 2022 at 2:11

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.
• 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? Commented Mar 7, 2022 at 16:40
• 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. Commented Mar 7, 2022 at 16:43
• @bamajon1974 The statement "$\lim_{x \to 2}\frac{1}{x^2} = L$" is identical to the statement "$\lim_{h \to 0}\frac{1}{(2 + h)^2} = L$". Commented Mar 7, 2022 at 19:59
• @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. Commented Mar 7, 2022 at 20:04