What does it mean to set $\delta$ to a minimum

calculusepsilon-deltalimitsproof-explanation

If we were to prove $\lim_{x \to 4} \sqrt{x} = 2$, then by using $\epsilon$–$\delta$ definition, we have:

$0 \leq |\sqrt{x} – 2| < \epsilon$

$-\epsilon< \sqrt{x} – 2 < \epsilon$

$2 -\epsilon< \sqrt{x} < 2 + \epsilon$

$(2 -\epsilon)^{2}< x < (2 + \epsilon)^{2}$

$4 -4\epsilon +\epsilon^{2}< x < 4 +4\epsilon -\epsilon^{2}$

Now, we set $\delta \leq \min(4\epsilon +\epsilon^{2}, 4\epsilon -\epsilon^{2})$ and choose whichever is smaller.

Since $\epsilon > 0$, the minimum is $\delta \leq 4\epsilon -\epsilon^{2}$

Source: Hartman, G. (n.d.). 1.2: Epsilon-Delta Definition of a Limit. LibreTexts. Retrieved April 29, 2021, from https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/01%3A_Limits/1.02%3A_Epsilon-Delta_Definition_of_a_Limit

Remark: I paraphrased what the author said, including only the key points, but the computations remain the same.

From the above, I have some questions, which I'll list as follows:

  1. The minimum, according to the source, is $\delta \leq 4\epsilon -\epsilon^{2}$. Is this because that's the only value $\delta$ that guarantees, for every $x$ within $\delta$ units from $4$, $f(x)$ is to be within $\epsilon$ units from 2?
  2. Does the need for minimum come from the fact that the distance between $4$ and $4 – \delta$ and the distance between $4$ and $4 + \delta$ is not the same?
  3. Is the distance between $4$ and $4 – \delta$ and the distance between $4$ and $4 + \delta$ not the same because the function we're evaluating is non-linear? Does this mean that the distances from $4$ are to be the same when the function is linear?
  4. Why can $\delta$ be made smaller than the minimum?

Please let me know if I've made myself understood as I'm not a native english speaker.

Best Answer

  1. The minimum, according to the source, is δ≤4ϵ−ϵ2. Is this because that's the only value δ that guarantees, for every x within δ units from 4, f(x) is to be within ϵ units from 2?

Because for any $\epsilon$, with that choose of $\delta$, $$ 4-\delta < x < 4+\delta \implies 2 - \epsilon < f(x) < 2 + \epsilon $$ And this satifies(actually, this is) the $\epsilon-\delta$ definition. So your statement is basically correct, the only problem is "the only value $\delta$ that satifies". Fix $\epsilon$, then with smaller choose of $\delta$, let's say some $\delta'$ < $\delta$, $$ 4-\delta' < x < 4+\delta' \implies 4-\delta < x < 4+\delta \implies 2 - \epsilon < f(x) < 2 + \epsilon $$ So in fact, any smaller choose of $\delta$ satifies the inequality.

  1. Does the need for minimum come from the fact that the distance between 4 and 4−δ and the distance between 4 and 4+δ is not the same?

I think you mean the distance between f(4) and f(4−δ) copmared to f(4) with f(4+δ). Yes. Because you need different choose of $\delta$ to satify both sides of the inequality, and you have to make the both sides correct to prove it satifies the definition.

  1. Is the distance between 4 and 4−δ and the distance between 4 and 4+δ not the same because the function we're evaluating is non-linear? Does this mean that the distances from 4 are to be the same when the function is linear?

I think you mean the distance between $f(4)$ and $f(4-δ)$ copmared to $f(4)$ with $f(4+δ)$. Yes, because we're evaluating a non-linear function. When the function is linear, it's clear that $$ f(4) - f(4-\delta) = f(4+\delta) - f(4)$$

  1. Why can δ be made smaller than the minimum?

Look 1.

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