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I'm looking for a verification of my $\epsilon-\delta$ proof of a limit example, if my proof is not completely mathematically rigorous, please tear it apart.


Required to Prove:

$$\lim_{x \to 1} \ \ \frac{2+4x}{3} = 2$$


Epsilon-Delta Definition

$$ \lim_{x \to a} f(x) = L \Leftrightarrow \forall \epsilon >0 (\exists \delta>0 \ni (\forall x \in D(0 < |x-a| < \delta \implies |f(x)-L| < \epsilon)))$$


Proof:


I start out with $0 < |x-a| < \delta$ and use it to prove that it implies $|f(x)-L| <\epsilon$

$$0 < |x-a| < \delta$$ $$Let \ \ a = 1$$ $$0 < |x-1| < \delta$$ Multiplying by 4 throughout the inequality we get $$4(0) < 4|x-1| < 4\delta$$ $$0 < |4x-4| < 4\delta$$ Diving by 3 throughout the inequality we get $$\frac{0}{3} < \frac{1}{3}|4x-4| < \frac{4}{3}\delta$$ $$0 < |\frac{4x-4}{3}| < \frac{4}{3}\delta$$ $$0 < |\frac{4x+2}{3}-2| < \frac{4}{3}\delta$$ Letting $f(x) = \frac{4x+2}{3}$ and $L=2$, we can see that we have something similar to what we want to arrive at $$0 <|f(x)-L| < \frac{4}{3}\delta$$ This suggests that we choose $\frac{4}{3}\delta = \epsilon \implies \delta = \frac{3}{4}\epsilon$ $$\implies |f(x)-L|<\epsilon \ , \ \ \ \forall \ \delta =\frac{3}{4}\epsilon$$ $$\implies \delta(\epsilon) = \frac{3}{4}\epsilon $$

Therefore we have proven that given any $\epsilon > 0$, we can always find a $\delta>0$, that satisfies the inequality $0 <|x-a|<\delta \implies |f(x)-L|<\epsilon$

$$ Q.E.D.$$


Some notes on my proof.

(and on writing delta as a function of epsilon)

Since $\delta$ is dependent on $\epsilon$ I've written $\delta$ as a function of $\epsilon$ , i.e. $\delta(\epsilon)$. Sometimes this is seen as a "no-no" when writing $\epsilon - \delta$ proofs, because as a previous user put it :

"I guess one could also write something like $\delta(\epsilon)$ although implying that delta is a function of epsilon is in general incorrect because we can usually find multiple delta satisfying the condition for a given epsilon, not just one." - User A

"If for a given $\epsilon$ a $\delta$ is "good", then any $0<\delta'<\delta$ is also "good". So I'm not sure it'd be a well-defined function, but you can certainly think of it as one - keeping in mind that there are a lot of other $\delta$'s for that $\epsilon$." - User B

This implies that $\delta(\epsilon)$ is not a function as for any input ($\epsilon$) we have multiple output ($\delta$'s)

But since we are only trying to prove the existence of a single $\delta$ (to satisfy the $\epsilon-\delta$ definition of a limit), shouldn't writing $\delta(\epsilon)$ still be considered mathematically rigorous given the fact that we are only finding a function $\delta(\epsilon)$ that has a co-domain of a subset of all possible $\delta$'s? If not what are possible ways to workaround this, and introduce further mathematical rigor to this last step?

So is my proof mathematically rigorous? Are there any errors I've made? If this proof is not completely mathematically rigorous, please tear it apart, I'm looking to improve my proof-writing skills as an undergraduate student majoring in Pure Mathematics, so please feel-free to offer any criticism and/or suggestions if you feel necessary.

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  • $\begingroup$ It seems to me you are right. The $\delta$ depends of $x$ excepting when the continuity is "uniform" (I don't know if you are aware of this notion stronger than simply continuity).Regards. $\endgroup$ – Piquito Apr 17 '16 at 20:02
  • $\begingroup$ There is nothing wrong with writing $\delta$ as a function of $\epsilon$. In fact, you have to do this with nearly all functions. The only exceptions I can think of are constant functions as well as some pathological examples like $f:(0,1) \to \mathbb{R}$ given by $f(x) = \begin{cases} 0 & x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{1}{q}, & x \in \mathbb{Q}, x=\frac pq\ \text{in lowest terms} \end{cases}$ For this, the $f(x) \to 0$ as $x \to a$ for any $a$ and our choice of $\delta$ is independent of $\epsilon$. $\endgroup$ – MathematicsStudent1122 Apr 17 '16 at 20:06
  • $\begingroup$ @MathematicsStudent1122 A commentator in a previous question I asked on $\epsilon-\delta$ proofs said the following about writing $\delta$ as a function of $\epsilon$ : "I guess one could also write something like $\delta(\epsilon)$ although implying that delta is a function of epsilon is in general incorrect because we can usually find multiple delta satisfying the condition for a given epsilon, not just one. " But since we only need to prove the existence of a single $\delta$ is it still not mathematically rigorous to write $\delta(\epsilon)$? $\endgroup$ – Perturbative Apr 17 '16 at 22:41
  • $\begingroup$ @Perturbative In general, if a particular $\delta$ works, then any $\lambda$ satisfying $0<\lambda <\delta$ also works, clearly. In that sense, no, the "function" $\delta(\epsilon)$ is not well defined. However, when we actually do these sorts of problems in practice, we merely want to find some delta and this delta is almost always, when we arrive at our final answer, expressed in terms of $\epsilon$. $\endgroup$ – MathematicsStudent1122 Apr 17 '16 at 22:49
  • $\begingroup$ @MathematicsStudent1122 I was thinking the same thing, but is there any way to introduce any further mathematical rigor to this last step? Is there another more rigorous way of concluding the proof than simply writing delta as a function of epsilon $\delta(\epsilon)$ and concluding from there? $\endgroup$ – Perturbative Apr 17 '16 at 22:52
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It appears that your proof is correct, albeit in a very verbose way. I would recommend thoroughly polishing it; of course, the ability to do that effectively comes with time: I hardly think I have mastered proof-writing, but I thought I would contribute what I think is a much more succinct proof of your claim.


Claim: $\displaystyle\lim_{x\to 1}\frac{2+4x}{3}=2$

Proof. Given $\epsilon>0$, choose $\delta=\frac{3}{4}\epsilon$. If $|x-1|<\delta$, then $$ \left|\frac{2+4x}{3}-2\right|=\frac{4}{3}|x-1|<\frac{4}{3}\delta=\frac{4}{3}\left(\frac{3}{4}\epsilon\right)=\epsilon. $$ Thus, if $|x-1|<\delta$, then $\left|\frac{2+4x}{3}-2\right|<\epsilon$. Therefore, by the definition of a limit, $\lim_{x\to 1}\frac{2+4x}{3}=2$, as desired. $\blacksquare$


Truthfully, the "Thus, if ..." part directly above is not even necessary. You could have concluded at the "... $=\epsilon$." part. Sometimes it simply comes down to a matter of style, but the rule of thumb is to not write more than is necessary. Your proof details more of all of your thought that went into it (and that is great in some contexts), but the final proof should be something much more clean and crisp. Hopefully you find that helpful in some manner.

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