0

My question builds on the well answered question from $\min$ in epsilon-delta and https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/01%3A_Limits/1.02%3A_Epsilon-Delta_Definition_of_a_Limit

In summary, we must prove $$\begin{align}\lim_{x \to 4} \sqrt{x} = 2\end {align}$$

In the steps following a delta is found such that $$\delta \leq min \{4\epsilon - {\epsilon }^{2}, 4\epsilon + {\epsilon}^{2}\}$$

The OP's original question involved why 'min' is used, and there is an extensive and accepted answer here https://math.stackexchange.com/a/3768851/827596

Now for my question.

If you look at the text where this example is taken from, you will see, almost as an after thought , the following.

Actually, it is a pain, but this won't work if ≥4 . This shouldn't really occur since is supposed to be small, but it could happen. In the cases where ≥4 , just take =1 and you'll be fine.

Now, I am not asking why an $\epsilon > 4$ will not work {it would produce a negative $\epsilon$}, my question is, hopefully, more nuanced. (A variation of this can be seen in many of these types of questions, where, almost inevitably, a commenter will point out that all the detailed 'scratch work' will fail, and a fall-back is needed for a larger than expected $\epsilon$)

I want to understand this 'afterthought' more fully. For example, it seems that the $\delta$s we so meticulously find satisfy a 'small' $\epsilon$, not a very large one.

The last issue I would like clarified, is this. The terminology “$\exists\delta$” in the formal definition of a limit, would seem, at first read, to imply that one $\delta$ exists for “$\forall \epsilon$s”, whereas it seems to be a more correct understanding that one may need more than one $\delta$ to prove the definition of a particular limit. (Which is I believe indirectly addressed by the answer below)

Thanks in advance.

user1115542
  • 523
  • The accepted answer here seems to address your question. – Cameron Buie Apr 25 '24 at 23:59
  • @CameronBuie thanks. That helps. – user1115542 Apr 26 '24 at 00:11
  • 1
    Well if $\epsilon_{big} > \epsilon_{small} > 0$ and $\delta$ works for $\epsilon_{small}$ then it will also work for $\epsilon_{big}$. So if we don't have a clearcut way of dealing with $\epsilon_{big}$ we can just use a solution for an $\epsilon_{small}$ and it will work. We could do a "If $\epsilon \ge 4$ let $\delta = 1$" or we could say "case 1: Assume $\epsilon < 4$" do a proof and "Case 2: $\epsilon \ge 4$. Find another $\epsilon_2 < 4 \le \epsilon$ and find the $\delta$ where $|x-a|<\delta\implies |f(x)-L|<\epsilon_2$. Then $|f(x)-L|<\epsilon_2 <4< \epsilon$; and we are done". – fleablood Apr 26 '24 at 15:43
  • @fleablood. Thanks. And I have enjoyed your other comments/answers in similar questions. – user1115542 Apr 26 '24 at 17:23
  • 1
    FWIW. "$\delta \leq min {4\epsilon - {\epsilon }^{2}, 4\epsilon + {\epsilon, }^{2}}$" is badly stated. $\epsilon^2 > 0$ so $min {4\epsilon - {\epsilon }^{2}, 4\epsilon + {\epsilon, }^{2}}= 4\epsilon -\epsilon^2$ always. And if $\epsilon \ge 4$ then $\delta < 0$ and isn't a valid choice. The thing is that the OP thought $4\epsilon + \epsilon^2$ would be valid. But they were wrong. – fleablood Apr 26 '24 at 17:23
  • 1
    "in the formal definition of a limit, would seem, at first read, to imply that one δ exists for “∀ϵs”" It does not mean that one $\delta$ exists. It means that at least one $\delta$ exists. And with a little thought we realize that if one $\delta_{prime}$ works then all $\delta: 0 < \delta < \delta_{prime}$ will also work because if $|x-a| < \delta < \delta_{prime}$ then $|x-a| < \delta_{prime}$. The terminology for exactly one $x$ exists would be $\exists! x$. $\exists x$ just means at least on exists. – fleablood Apr 26 '24 at 17:27
  • @fleablood Thanks for pointing out that subtlety. It has taken me quite a while to realize what that really implies. – user1115542 Apr 26 '24 at 17:37

1 Answers1

0

The definition of $\lim_{x \rightarrow 4} \sqrt{x} = 2$ is that for all $\varepsilon > 0$, there exists $\delta > 0$ such that if $|x-4| < \delta$, then $|\sqrt{x}-2|<\varepsilon$.

This limit is proven in the example by giving the explicit expression $\delta = \min\{4\varepsilon - \varepsilon^2,4\varepsilon + \varepsilon^2\}$. The trouble is that we must show that for all $\varepsilon > 0$, there exists $\delta > 0$ such that if $|x-4| < \delta$, then $|\sqrt{x}-2|<\varepsilon$. If $\varepsilon \ge 4$, then that expression for $\delta$ is non-positive. One could then argue that you have only shown this statement holds for all $0 < \varepsilon < 4$.

In practice however, only small $\varepsilon$ are troublesome and often large $\varepsilon$ are treated as trivial. The author proposes a complete expression for $\delta$ as

$$\delta = \begin{cases} \min\{4\varepsilon - \varepsilon^2,4\varepsilon + \varepsilon^2\} & \varepsilon < 4 , \\ 1 & \varepsilon \ge 4, \end{cases}$$

which will work for all $\varepsilon > 0$.

Brian Lai
  • 850
  • The last part of your answer is what my thoughts have begun to align with. Thanks. – user1115542 Apr 26 '24 at 00:07
  • 1
    It might be worth working through $\delta =1$. $|x-4| < 1 \implies 3 < x < 5\implies \sqrt 3 < \sqrt x < \sqrt 5\implies -2 < \sqrt 3 < \sqrt x <\sqrt 5 < 6$. Now as $\epsilon \ge 4$ we have $2-\epsilon \le -2 < \sqrt x < 6\le 2+\epsilon$ so $|\sqrt x -2| <\epsilon$. To my mind this is overkill, and misses the important concept: If you can prove something for a small $\epsilon$ that that we prove it also for every bigger $\epsilon$ so it is always good enough to prove it for an $\epsilon$ that we assume is "small enough". – fleablood Apr 26 '24 at 17:35
  • @fleablood. Nice…and thank you again. I think I got it. :-) – user1115542 Apr 26 '24 at 17:42